Thermodynamics - Class 11 Physics - Chapter 11 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Thermodynamics | NCERT | Physics | Class 11
The criteria for spontaneity among the following is
A $\Delta \mathrm{S}_{\text{system}} + \Delta \mathrm{S}_{\text{surrounding}} < 0$
B $\mathrm{dU} < \mathrm{O}$ at constant $\mathrm{S}$ and $\mathrm{V}$
C $\mathrm{dH} < \mathrm{O}$ at constant $\mathrm{S}$ and $\mathrm{P}$
D $\mathrm{dH} < \mathrm{O}$ at constant $\mathrm{S}$ and $\mathrm{T}$
The correct options indicating the criteria for spontaneity are: Option B: $\mathrm{dU} < 0$ at constant $S$ and $V$ Option C: $\mathrm{dH} < 0$ at constant $S$ and $P$
Explanation:
Option A: For a spontaneous process, the total entropy change, which includes both the system and surroundings, must be positive: $$ \Delta \mathrm{S}_{\text{system}} + \Delta \mathrm{S}_{\text{surroundings}} > 0 $$ Therefore, option A is incorrect as it states the opposite.
Option B: At constant entropy ($S$) and volume ($V$), $\mathrm{dU}$ corresponds to the change in internal energy under these constraints. According to the Gibbs free energy equation $$ \Delta G = \Delta H - T \Delta S $$ and at constant $S$, $\Delta S = 0$, thus $\mathrm{dG} = \mathrm{dU}$. For spontaneity, $\mathrm{dG}$ must be negative which implies $\mathrm{dU} < 0$.
Option C: Similarly, at constant entropy and pressure, $\mathrm{dG} = \mathrm{dH}$. For a process to be spontaneous, $\mathrm{dG}$ needs to be negative, hence $\mathrm{dH}$ should also be negative under these conditions.
Option D: At constant entropy and temperature, changing enthalpy does not affect Gibbs free energy directly since entropy is constant and temperature is a consistent factor; thus, $\mathrm{dH}$ alone does not directly dictate spontaneity. Typically, $\mathrm{dG}$ at constant $S$ and $T$ can actually be zero based on equilibrium conditions rather than indicating spontaneity through being negative.
In conclusion, the criteria for spontaneous changes are correctly given in Options B and C. As per these conditions, internal energy and enthalpy must decrease under specific constant conditions to drive a process spontaneously.
Which of the following statements is correct with respect to ocean thermal energy?
A. It produces a lot of pollution.
B. It can destroy life in the ocean.
C. It uses the temperature difference in the ocean to generate energy.
D. None of these
The correct answer is C. It uses the temperature difference in the ocean to generate energy.
Ocean Thermal Energy Conversion (OTEC) is a type of non-conventional energy that harnesses the temperature difference between the warmer surface water and the much cooler deep ocean water to produce electricity. This system primarily utilizes the thermal gradient which can be about 40 degrees different in tropical regions.
OTEC technologies leverage this significant temperature difference typically around 1000 meters deep, to operate power-generating cycles. Notably, it is considered that OTEC could potentially produce more energy than other conventional marine energy sources like tidal, wave, or wind energy. Furthermore, it is environmentally friendly since it does not produce pollution and does not harm marine life, making it a sustainable and renewable source of energy.
For the reaction $\mathrm{CO}(\mathrm{g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})$, $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ are $-283$ kJ and $-87$ J K$^{-1}$ mol$^{-1}$, respectively. It was intended to carry out this reaction at 1000, 1500, 3000, and 3500 K. At which of these temperatures would this reaction be thermodynamically spontaneous?
A. 1500 and 3500 K
B. 3000 and 3500 K
C. 1000, 1500, and 3000 K
D. 1500, 3000, and 3500 K
To determine at which temperatures the reaction $$\mathrm{CO}(\mathrm{g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})$$ is thermodynamically spontaneous, we use the relationship:
$$ \Delta G = \Delta H - T\Delta S $$
Where:
$\Delta G$ is the Gibbs free energy change,
$\Delta H$ is the enthalpy change,
$\Delta S$ is the entropy change,
$T$ is the temperature in Kelvin.
Given values: $\Delta H = -283$ kJ and $\Delta S = -87$ J K$^{-1}$ mol$^{-1}$.
Calculations:
At $1000 \text{ K}$:$$ \Delta G = -283, \text{kJ} - 1000 \text{ K} \times (-87 \times 10^{-3}, \text{kJ K}^{-1}\text{mol}^{-1}) $$ $$ \Delta G = -283, \text{kJ} + 87, \text{kJ} = -196, \text{kJ} $$ Result: $\Delta G = -196$ kJ, spontaneous.
At $1500 \text{ K}$:$$ \Delta G = -283, \text{kJ} + 1500 \times 0.087, \text{kJ} = -283, \text{kJ} + 130.5, \text{kJ} = -152.5, \text{kJ} $$ Result: $\Delta G = -152.5$ kJ, spontaneous.
At $3000 \text{ K}$:$$ \Delta G = -283, \text{kJ} + 3000 \times 0.087, \text{kJ} = -283, \text{kJ} + 261, \text{kJ} = -22, \text{kJ} $$ Result: $\Delta G = -22$ kJ, spontaneous.
At $3500 \text{ K}$:$$ \Delta G = -283, \text{kJ} + 3500 \times 0.087, \text{kJ} = -283, \text{kJ} + 304.5, \text{kJ} = 21.5, \text{kJ} $$ Result: $\Delta G = 21.5$ kJ, non-spontaneous.
Conclusion:
The reaction is thermodynamically spontaneous at 1000 K, 1500 K, and 3000 K. Therefore, the correct option is C.
At constant pressure ($\Delta P = 0$), then enthalpy is: $$ \begin{array}{l} \Delta H = \Delta U + \Delta(P V) \ \Delta H = \Delta U + P \Delta V + V \Delta P; \text{ but as } (\Delta P = 0) \ \Delta H = \Delta U + P \Delta V \end{array} $$
But if the volume is constant ($\Delta V = 0$), then the equation becomes: $$ \Delta H = \Delta U + V \Delta P $$
What does this equation imply and what are its applications?
The given equations help to estimate the change in enthalpy ($\Delta H$) under different conditions.
If the pressure is constant ($\Delta P = 0$), the equation simplifies to: $$ \Delta H = \Delta U + P \Delta V $$ This equation is valuable in scenarios where pressure is kept constant, such as in many chemical reactions and process engineering operations. It allows us to calculate the change in enthalpy by considering the internal energy change ($\Delta U$) and the change in volume ($\Delta V$) multiplied by the constant pressure ($P$).
Conversely, if the volume is constant ($\Delta V = 0$), the equation becomes: $$ \Delta H = \Delta U + V \Delta P $$ This formulation becomes practical where volume remains unchanged, such as in isochoric processes often observed in closed systems like bomb calorimeters used for combustion studies.
Hence, these equations are directly applicable in calculating the change in enthalpy under specific conditions of constant pressure or constant volume, simply by identifying these conditions from the problem statement.
A sample of ideal gas ($\gamma=1.4$) is heated at constant pressure. If $140 \mathrm{~J}$ of heat is supplied to the gas, find $\Delta U$ and $w$.
A) $60 \mathrm{~J}, -80 \mathrm{~J}$
B) $50 \mathrm{~J}, -60 \mathrm{~J}$
C) $-50 \mathrm{~J}, 60 \mathrm{~J}$
D) None of these
The correct option is $\mathbf{A}$: $$ 60 \mathrm{~J}, -80 \mathrm{~J} $$
Remember the specific heat ratio $ \gamma $ for an ideal gas: $$ \gamma = 1.4 $$ which implies: $$ \frac{C_p}{C_V} = 1.4 \Rightarrow C_p = 1.4 C_V $$
From the relation $C_P - C_V = R$: $$ \begin{align*} 1.4 C_V - C_V &= R \ 0.4 C_V &= R \ \therefore C_V &= \frac{R}{0.4} = \frac{5}{2} R \end{align*} $$ and similarly: $$ \begin{align*} C_P &= C_V + R = \frac{5}{2} R + R = \frac{7}{2} R \end{align*} $$
The heat transferred at constant pressure $Q_P$ equals the change in enthalpy $\Delta H$: $$ \Delta H = Q_P = n C_P \Delta T $$ Rearranging for $\Delta T$: $$ \Delta T = \frac{Q_P}{n C_P} = \frac{140}{n \cdot \frac{7}{2} R} = \frac{40}{n} $$
Work done $w$ is given by: $$ w = -n R \Delta T = -n \cdot 2 \cdot \frac{40}{n} = -80 \mathrm{~J} $$ Hence, $w = -80 \mathrm{~J}$.
The change in internal energy $\Delta U$ can also be calculated using: $$ Q_P = \Delta H = \Delta U + (-w) $$ $$ \Delta U = \Delta H + w = 140 + (-80 \mathrm{~J}) = 60 \mathrm{~J} $$
Thus, $$ \Delta U = 60 \mathrm{~J} \quad \text{and} \quad w = -80 \mathrm{~J} $$
Conclusion: The correct answers are $\Delta U = 60 \mathrm{~J}$ and $w = -80 \mathrm{~J}$.
n moles of a gas filled in a container at temperature $T$ is in thermodynamic equilibrium initially. If the gas is compressed slowly and isothermally to half its initial volume, then the work done by the atmosphere on the piston is -
A $nRT \ln 2$
B $\frac{nRT}{2}$
C $-\frac{3nRT}{2}$ D $\frac{3nRT \ln 2}{2}$
The correct answer is Option B $$\frac{nRT}{2}$$.
To determine the work done by the atmosphere on the piston during this isothermal compression of the gas, we analyze the process as follows:
Since the compression is isothermal (temperature $T$ remains constant) and slow, it allows the gas to maintain equilibrium with the surrounding atmosphere at pressure $P_{\text{atm}}$ throughout the process.
Work done by the atmosphere on the piston can be computed using the formula: $$ W = P_{\text{atm}} \Delta V = P_{\text{atm}} (V_{i} - V_{f}) $$ Given that the final volume $V_{f}$ is half the initial volume $V_{i}$, we have: $$ V_{f} = \frac{V_{i}}{2} $$ Therefore, the change in volume $\Delta V$ is: $$ \Delta V = V_{i} - \frac{V_{i}}{2} = \frac{V_{i}}{2} $$ Substituting back, we find: $$ W = P_{\text{atm}} \frac{V_{i}}{2} $$ Using the ideal gas law initially: $$ P_{\text{gas}} V_{i} = nRT \rightarrow P_{\text{atm}} V_{i} = nRT $$ since initially $P_{\text{gas}} = P_{\text{atm}}$. Replacing in the formula for work done, we obtain: $$ W = \frac{nRT}{2} $$
Thus, the work done by the atmosphere on the piston is $$\frac{nRT}{2}$$, corresponding to Option B.
If for a gaseous reaction, ΔH is the enthalpy change and ΔU is the internal energy change, then:
ΔH is always greater than ΔU.
ΔH < ΔU if the number of moles of products is greater than the number of moles of reactants.
ΔH is always less than ΔU.
ΔH < ΔU if the number of moles of products is greater than the number of moles of reactants.
Correct Answer: D
If $\Delta n$ is negative, then $\Delta H < \Delta U$.
For the hypothetical reaction $ A \rightarrow B $, the activation energies for the forward and reverse reactions are 19 kJ/mol and 9 kJ/mol, respectively. The value of the reaction enthalpy is:
A. 28 kJ
B. 19 kJ
C. 10 kJ
D. 9 kJ
For a hypothetical reaction $ A \rightarrow B $, the activation energy for the forward and reverse reactions are given as 19 kJ/mol and 9 kJ/mol respectively. To find the enthalpy change ($ \Delta H $) of the reaction, we use the formula:
$$ \Delta H = E_a (\text{forward}) - E_a (\text{reverse}) $$
Given:
Activation energy for the forward reaction ($ E_a (\text{forward}) $) = 19 kJ/mol
Activation energy for the reverse reaction ($ E_a (\text{reverse}) $) = 9 kJ/mol
Thus,
$$ \Delta H = 19 , \text{kJ/mol} - 9 , \text{kJ/mol} = 10 , \text{kJ/mol} $$
Therefore, the enthalpy change for the reaction is 10 kJ/mol.
Final Answer: C
The internal energy of a substance:
A: Increases when the temperature is increased.
B: Decreases when the temperature is increased.
C: Can be calculated using $ E = mc^{2} $.
D: Remains unaffected by temperature changes.
The correct answer is: A: It increases with increasing temperature.
Explanation: The internal energy ($ \Delta U $) of a substance increases with the increase in temperature.
Final Answer: A
A body reaches state A from B through one path and reaches state B from A through another path. If the corresponding internal energy changes for these paths are $\Delta U_1$ and $\Delta U_2$ respectively, then:
A. $ U_1 + U_2 = - ve $
B. $ U_1 + U_2 = + ve $
C. $ U_1 + U_2 = 0 $
D. None of the above
The correct answer is: C
For a cyclic process, the change in internal energy, $$ \Delta U $$, is always zero.
Therefore, $$ U_1 + U_2 = 0 $$.
Final Answer: C
The heat of combustion of carbon monoxide at constant volume and $17^\circ C$ is 283.3 kilojoules. The value of the heat of combustion at constant pressure is: (R = 8.314 joules degree $^{-1}$ mole $^{-1}$)
A) -284.5 kilojoules B) 284.5 kilojoules C) 384.5 kilojoules D) -384.5 kilojoules
To determine the enthalpy change ($\Delta H$) at constant pressure given the internal energy change ($\Delta U$) at constant volume, we can use the following process:
Start with the combustion reaction for carbon monoxide:
$$ \mathrm{CO}_{(g)} + \frac{1}{2} \mathrm{O_{2(g)}} \rightarrow \mathrm{CO_{2(g)}} $$
Use the relationship that connects $\Delta H$ and $\Delta U$:
$$ \Delta H = \Delta U + \Delta n \cdot R \cdot T $$
Calculate the change in moles of gas ($\Delta n$):
$$ \Delta n = (\text{moles of gas products}) - (\text{moles of gas reactants}) $$ $$ \Delta n = 1 - \left( 1 + \frac{1}{2} \right) = -\frac{1}{2} $$
Given data:
$\Delta U = -283.3 \times 10^{3}$ J (since 283.3 kJ is converted to J)
$R = 8.314$ J/(mol·K)
$T = 17^\circ C = 290$ K (Convert Celsius to Kelvin by adding 273)
Substitute the values into the equation:
$$ \Delta H = -283.3 \times 10^{3} + \left( -\frac{1}{2} \right)(8.314 \times 290) $$
Perform the multiplication and addition:
$$ \Delta H = -283.3 \times 10^{3} - \frac{1}{2}(8.314 \times 290) $$ $$ \Delta H = -283300 - 1.206(290) $$ $$ \Delta H = -283300 - 1205.53 $$ $$ \Delta H = -284505.53 \text{ J} = -284.5 \text{ kJ} $$
Therefore, the enthalpy change at constant pressure is -284.5 kJ.
Final Answer: A
In thermodynamics, a process is called reversible when:
A. The surroundings and the system can be transformed into one another.
B. There is no boundary between the system and the surroundings.
C. The surroundings and the system are always in equilibrium with each other.
D. The system spontaneously transforms into the surroundings.
The correct answers are:
A, C
In thermodynamics, a reversible process is one in which the surroundings and the system are always in equilibrium with each other.
The possible enthalpies of compounds $x$ and $y$ are -84 kJ and -156 kJ, respectively. Which of the following statements is correct with reference to the above?
A. $x$ is more stable than $y$.
B. $x$ is less stable than $y$.
C. Both $x$ and $y$ are unstable.
D. $x$ and $y$ are endothermic compounds.
Correct Answer: B
Explanation: Since the energy of $y$ is less than that of $x$, it means that $y$ is more stable than $x$.
Final Answer: B
Which of the following is not an intensive property in thermodynamics?
A) Pressure
B) Density
C) Volume
D) Temperature
Answer: C
Volume is not an intensive property.
In a closed, thermally insulated vessel, a liquid is stirred by a paddle to increase its temperature. Which of the following statements is true for this process?
A. $ \Delta U = w = Q = 0 $
B. $ \Delta U \ne 0, Q = w = 0 $
C. $ \Delta U = w \ne 0, Q = 0 $
D. $ \Delta U = Q \ne 0, w = 0 $
In a closed and insulated system, heat transfer, $ \mathrm{q} $, is zero. Therefore, the change in internal energy, $ \Delta U $, is given by the work done on the system. Expressing this mathematically:
$$ \Delta U = q + w $$
Since $ q = 0 $ for an insulated system:
$$ \Delta U = w $$
Thus, we can summarize:
$$ \Delta U \neq 0, \ w \neq 0, \ q = 0 $$
Therefore, the correct statement is C: $ \Delta U = w \neq 0, \ q = 0 $.
What do you understand by the term ‘Specific Latent Heat of Fusion of iron is 40000 Jkg-1’.
A. 40000 J of heat is required to melt 1 kg of iron at its melting point without any change in its temperature
B. 40000 J of heat is required to melt 1 kg of iron at any temperature
C. 40000 J of heat is required to melt 1 g of iron at any temperature
D. 40000 J of heat is required to melt 1 g of iron at its melting point without any change in its temperature
E. 40000 J of heat is liberated in melting 1 kg of iron at its melting point without any change in its temperature
The correct option is A:
40000 J of heat is required to melt 1 kg of iron at its melting point without any change in its temperature.
To melt 1 kg of iron at its melting point without any change in temperature, 40000 J of heat energy is required.
If the internal energy of an ideal gas decreases by the same amount as the work done by the system, the process is:
A. Cyclic
B. Isothermal
C. Adiabatic
D. Isolated
The correct option is $\mathbf{C}$ Adiabatic.
In this scenario, the internal energy of the ideal gas decreases by the same amount as the work done by the system.
Referring to the first law of thermodynamics, which states:
$$ \Delta U = q + w $$
where:
$\Delta U$ is the change in internal energy
$q$ is the heat exchange
$w$ is the work done
In an adiabatic process, there is no heat exchange with the surroundings, meaning $q = 0$. Therefore, the equation simplifies to:
$$ \Delta U = w $$
This indicates that the change in internal energy is solely due to the work done by or on the system. As a result, when the internal energy decreases by the same amount as the work done by the system, it signifies an adiabatic process where a decrease in temperature occurs due to the absence of any heat exchange.
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A geyser heats water flowing at the rate of 3.0 litres per minute from $27^{\circ} \mathrm{C}$ to $77^{\circ} \mathrm{C}$. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is $4.0 \times 10^{4} \mathrm{~J} / \mathrm{g}$ ?
Rate of Fuel Consumption
The rate of fuel consumption is approximately 15.7 grams per minute.
Thus, to maintain the heating, the rate of fuel consumption is 15.7 g/min when the geyser heats water flowing at the rate of 3.0 litres per minute from $27^{\circ} \mathrm{C}$ to $77^{\circ} \mathrm{C}$.
What amount of heat must be supplied to $2.0 \times 10^{-2} \mathrm{~kg}$ of nitrogen (at room temperature) to raise its temperature by $45{ }^{\circ} \mathrm{C}$ at constant pressure ? (Molecular mass of $\mathrm{N}_{2}=28 ; R=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$.)
The calculations are as follows:
Number of moles of nitrogen ($ n $):[ n = \frac{2.0 \times 10^{-2} , \text{kg}}{0.028 , \text{kg/mol}} = 0.714 , \text{mol} ]
Molar specific heat capacity at constant pressure ($C_p $):[ C_p = \frac{7}{2} R = \frac{7}{2} \times 8.3 , \text{J/mol⋅K} = 29.05 , \text{J/mol⋅K} ]
Amount of heat required ($ \Delta Q $):[ \Delta Q = n C_p \Delta T = 0.714 , \text{mol} \times 29.05 , \text{J/mol⋅K} \times 45 , \text{K} = 934 , \text{J} ]
Thus, the amount of heat that must be supplied to $2.0 \times 10^{-2} , \text{kg}$ of nitrogen to raise its temperature by $45{ }^{\circ} \mathrm{C}$ at constant pressure is 934 J.
Explain why
(a) Two bodies at different temperatures $T_{1}$ and $T_{2}$ if brought in thermal contact do not necessarily settle to the mean temperature $\left(T_{1}+T_{2}\right) / 2$.
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
(a) Two bodies at different temperatures $ T_{1} $ and $ T_{2} $ if brought in thermal contact do not necessarily settle to the mean temperature $\left(T_{1} + T_{2}\right) / 2 $.
When two bodies at different temperatures are brought into thermal contact, thermal equilibrium is reached when both bodies attain the same temperature. However, this equilibrium temperature depends on the specific heat capacities and masses of the two bodies, not simply their initial temperatures.
If the two bodies have vastly different heat capacities, the final temperature will be closer to the initial temperature of the body with the larger heat capacity. The formula for the final equilibrium temperature $ T_f $ can be derived from the principle of conservation of energy:
[ m_1 c_1 \left( T_f - T_1 \right) + m_2 c_2 \left( T_f - T_2 \right) = 0 ]
Solving the above equation gives the equilibrium temperature, which is generally not the arithmetic mean.
(b) The coolant in a chemical or nuclear plant should have high specific heat.
Specific heat capacity measures how much heat energy a substance can absorb per unit mass before its temperature rises significantly. A coolant with a high specific heat capacity can absorb large amounts of heat energy with minimal temperature increase. This is crucial in maintaining the machinery's components at a safe operating temperature, thereby preventing overheating and potential damage.
(c) Air pressure in a car tyre increases during driving.
During driving, the friction between the tyres and the road surface generates heat. Additionally, the repeated flexing of the tyre rubber as it rotates also contributes to increasing its temperature. According to the ideal gas law,$ PV = nRT $, where an increase in temperature ( T ) leads to an increase in pressure ( P ) if the volume ( V ) of the tyre and the amount of gas ( n ) remain constant. Hence, as the tyre heats up, the air inside expands, increasing the pressure.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
Harbour towns are located near large bodies of water which have a high specific heat capacity. Water heats up and cools down more slowly than land. As a result, the large water body acts as a heat reservoir:
In summer:The water absorbs heat without a significant rise in temperature, making coastal areas cooler.
In winter:The water releases the absorbed heat, keeping the coastal areas warmer.
This moderating effect leads to a more temperate climate in harbour towns compared to towns in deserts, where the temperature can vary drastically between day and night due to the absence of such a moderating influence.
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?
The factor by which the pressure increases is:
[ \left(\frac{1}{2}\right)^{-1.4} \approx 2.639 ]
So, the pressure of the gas increases by a factor of approximately 2.64 when the gas is compressed to half its original volume.
In changing the state of a gas adiabatically from an equilibrium state $A$ to another equilibrium state $B$, an amount of work equal to $22.3 \mathrm{~J}$ is done on the system. If the gas is taken from state $A$ to $B$ via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case ? (Take $1 \mathrm{cal}=4.19 \mathrm{~J}$ )
To determine the net work done by the system when it is taken from state $A$ to state $B$ via a process where the net heat absorbed is 9.35 cal, we can use the first law of thermodynamics.
The first law of thermodynamics is given by:
$$ \Delta Q = \Delta U + \Delta W $$
where:
$\Delta Q$ is the net heat absorbed by the system,
$\Delta U$ is the change in internal energy of the system,
$\Delta W$ is the net work done by the system.
For an adiabatic process, $\Delta Q = 0$, so the change in internal energy $\Delta U$ is equal to the work done on the system:
$$ \Delta U = -\Delta W_{adiabatic} $$
Given that the work done on the system adiabatically is 22.3 J, we have:
$$ \Delta U = -22.3 , \text{J} $$
Next, when the gas is taken from state $A$ to state $B$ via a process absorbing 9.35 cal of heat, we convert this heat into joules:
$$ \Delta Q = 9.35 , \text{cal} \times 4.19 , \text{J/cal} = 39.1865 , \text{J} $$
Using the first law of thermodynamics, we now solve for $\Delta W$:
$$ \Delta W = \Delta Q - \Delta U $$
Substitute the values:
$$ \Delta W = 39.1865 , \text{J} - (-22.3 , \text{J}) $$
$$ \Delta W = 39.1865 , \text{J} + 22.3 , \text{J} $$
$$ \Delta W = 61.4865 , \text{J} $$
Therefore, the net work done by the system in the latter process is 61.49 J (to two decimal places).
Two cylinders $A$ and $B$ of equal capacity are connected to each other via a stopcock. $A$ contains a gas at standard temperature and pressure. $B$ is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following :
(a) What is the final pressure of the gas in $A$ and $B$ ?
(b) What is the change in internal energy of the gas ?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its $P-V-T$ surface ?
Part (a)
Final Pressure of the Gas in $A$ and $B$
Initially, cylinder $A$ contains the gas at standard temperature and pressure, which means:
$P_A$ (initial pressure in $A$) = $1$ atm (standard pressure)
Volume of each cylinder $V = V_A = V_B$
Since cylinder $B$ is evacuated, $P_B$ (initial pressure in $B$) = $0$ atm.
When the stopcock is opened, the gas will expand to fill both cylinders equally. According to Boyle's Law ($P_1V_1 = P_2V_2$ for a constant temperature), the final pressure will be halved because the volume has doubled.
The total volume after opening the stopcock is $2V$. Therefore, the final pressure, $P_f$, can be calculated as:
$$ P_f = \frac{P_A \cdot V_A}{V_A + V_B} = \frac{1 \text{ atm} \cdot V}{2V} = \frac{1}{2} \text{ atm} = 0.5 \text{ atm} $$
So, the final pressure $P_f$ is 0.5 atm in both $A$ and $B$.
Part (b)
Change in Internal Energy of the Gas
Since the entire system is thermally insulated and the gas expands into a vacuum, no heat is exchanged (adiabatic process) and no work is done (free expansion). For an ideal gas undergoing a free (adiabatic) expansion, the change in internal energy $\Delta U$ depends solely on the temperature change:
$$ \Delta U = nC_V \Delta T $$
However, since there is no temperature change in an ideal gas during free expansion, the change in internal energy $\Delta U$ is 0.
Part (c)
Change in the Temperature of the Gas
In free expansion of an ideal gas, there's no work done and no heat exchanged in the ideal case:
$$ \Delta T = 0 $$
Therefore, the change in temperature of the gas is 0; the final temperature remains the same as the initial temperature.
Part (d)
Do the Intermediate States of the System (before Settling to the Final Equilibrium State) Lie on its $P-V-T$ Surface?
The intermediate states during the sudden opening of the stopcock do not lie on the $P-V-T$ surface. This is because the process is highly non-equilibrium: the gas rapidly expands into a vacuum without any quasi-static steps. During this rapid expansion, well-defined macroscopic variables like pressure and temperature cannot be assigned uniformly throughout the system. Therefore, the intermediate states before the system reaches equilibrium are not described by the thermodynamic $P-V-T$ surface.
Summary:
Final pressure in $A$ and $B$: 0.5 atm
Change in internal energy: 0 (zero)
Change in temperature: 0 (zero)
Intermediate states lie on the $P-V-T$ surface: No
An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
To determine the rate at which the internal energy of the system is increasing, we can use the First Law of Thermodynamics:
$$ \Delta Q = \Delta U + \Delta W $$
Here:
(\Delta Q) is the power supplied to the system by the heater (100 W)
(\Delta W) is the power the system uses to perform work (75 J/s or 75 W)
(\Delta U) is the rate at which the internal energy is increasing ((U))
We rearrange the equation to solve for (\Delta U):
$$ \Delta U = \Delta Q - \Delta W $$
Let’s plug in the values:
$$ \Delta U = 100\ W - 75\ W $$
So, the rate at which the internal energy is increasing is:
$$ \Delta U = 25\ W $$
Therefore, the internal energy of the system is increasing at a rate of 25 W.
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (11.13)
Fig. 11.11
Its volume is then reduced to the original value from $\mathrm{E}$ to $\mathrm{F}$ by an isobaric process. Calculate the total work done by the gas from $\mathrm{D}$ to $\mathrm{E}$ to $\mathrm{F}$
To calculate the total work done by the gas as it undergoes the change from $\text{D} \rightarrow \text{E} \rightarrow \text{F} $, we need to find the work done during each segment of the process and sum them up.
1. Work Done During $ \text{D} \rightarrow \text{E} $
The path from $ \text{D} $ to $\text{E} $ is a linear process. We need to approximate this as a combination of an isobaric and an isochoric process or directly integrate to find the work done.
For a linear process represented as a straight line on a PV-diagram, the work done can be found using the area under the process curve.
2. Work Done During $ \text{E} \rightarrow \text{F} $
This path is isobaric, i.e., the pressure is constant. The work done in an isobaric process is given by: [ W = P \Delta V ]
We will utilize these two pieces of information to find the total work done.
Calculation Steps
Work Done from $\text{D} \rightarrow \text{E} $
The work done can be represented as the area under the curve in the PV-diagram.
Since (\text{D} \rightarrow \text{E} ) is a straight line: [ \text{D} (V_1, P_1) \rightarrow \text{E} (V_2, P_2) ] where: $ V_1 = 2 , m^3 $ $V_2 = 5 , m^3 $ $ P_1 = 600 , N/m^2 $ $ P_2 = 300 , N/m^2$
The area under the curve can be calculated as the trapezoid area: [ A_{\text{trapezoid}} = \frac{1}{2}(P_1 + P_2)(V_2 - V_1) ]
Work Done from ( \text{E} \rightarrow \text{F} )
For the isobaric process: [ P = 300 , N/m^2 ] [ \Delta V = V_{\text{final}} - V_{\text{initial}} = 2 , m^3 - 5 , m^3 = -3 , m^3 ]
The work done is: [ W_{\text{E} \rightarrow \text{F}} = P \Delta V = 300 , N/m^2 \times (-3 , m^3) = -900 , J ]
Now let's calculate these explicitly.
Step-by-Step Calculation Using Wolfram Alpha
We'll perform the step-by-step computation to get the exact figures.
Step 1: Area under the curve $\text{D} \rightarrow \text{E}$
[ A_{\text{trapezoid}} = \frac{1}{2}(600 + 300)(5 - 2) = \frac{1}{2}(900)(3) = 1350 , J ]
Step 2: Work Done $\text{E} \rightarrow \text{F}$
[ W_{\text{E} \rightarrow \text{F}} = 300 \times (-3) = -900 , J ]
Total Work Done
[ W_{\text{total}} = W_{\text{D} \rightarrow \text{E}} + W_{\text{E} \rightarrow \text{F}} = 1350 - 900 = 450 , J ]
Conclusion
The total work done by the gas as it undergoes the changes from $ \text{D} \rightarrow \text{E} \rightarrow \text{F} $is: [ \boxed{450 , J} ]
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Comprehensive Thermodynamics Class 11 Notes: Key Concepts and Formulas
Introduction to Thermodynamics
Definition and Importance
Thermodynamics is the branch of physics that deals with the concepts of heat, temperature, and the inter-conversion of heat and other forms of energy. In Class 11, thermodynamics plays a crucial role in understanding various physical phenomena, including engines, refrigerators, and even the biological processes within the human body.
Historical Development of Heat Concept
Historically, heat was once thought to be an invisible fluid known as "caloric." This concept was later discarded in favour of modern thermodynamics, which views heat as a form of energy. Count Rumford's experiments in 1798 significantly contributed to this shift by demonstrating the conversion of mechanical work into heat.
Key Terms: Heat, Work, Temperature
In thermodynamics, heat is the transfer of energy due to temperature differences, work is the energy transfer associated with force acting through a distance, and temperature is a measure of the average kinetic energy of the molecules within a system.
Laws of Thermodynamics
The Zeroth Law of Thermodynamics
Statement and Explanation
The Zeroth Law states that if two systems (A and B) are each in thermal equilibrium with a third system (C), they are in thermal equilibrium with each other. This law forms the basis for the definition of temperature.
Concept of Temperature
Temperature, as inferred from the Zeroth Law, is a physical quantity that remains constant for two systems in thermal equilibrium. It acts as a measure of the 'hotness' or 'coldness' of a body.
Thermal Equilibrium
Thermal equilibrium is the state in which there is no net flow of heat between systems; their temperatures are equal.
The First Law of Thermodynamics
Statement and Explanation
The First Law of Thermodynamics states that the total energy of an isolated system is constant. It is essentially the law of conservation of energy applied to thermodynamic systems. [ \Delta Q = \Delta U + \Delta W ] Where:
$\Delta Q$ is the heat supplied to the system,
$\Delta U$ is the change in internal energy,
$\Delta W$ is the work done by the system.
Internal Energy
Internal energy ($U$) is the total energy of molecules within a system, including both kinetic and potential energies.
Work and its Calculations
Work in thermodynamics is often calculated through volume changes. For an ideal gas, work done is given by: [ \Delta W = P \Delta V ]
Examples and Applications
Practical applications of the First Law include the functioning of internal combustion engines and the assessment of energy changes in chemical reactions.
The Second Law of Thermodynamics
Statement and Explanation
The Second Law of Thermodynamics introduces the concept of entropy and states that in any energy transfer or transformation, the total entropy of an isolated system can only increase. [ \eta = 1 - \frac{T_2}{T_1} ] Where:
$\eta$ is the efficiency,
$T_1$ and $T_2$ are the temperatures of the hot and cold reservoirs.
Kelvin-Planck and Clausius Statements
Kelvin-Planck Statement: No process is possible where heat is completely converted into work.
Clausius Statement: No process is possible where heat flows from a colder object to a hotter object without external work.
Implications on Heat Engines and Refrigerators
The Second Law limits the efficiency of heat engines and the performance of refrigerators, asserting that no engine can be 100% efficient.
Thermodynamic Processes
Types of Thermodynamic Processes
Isothermal Processes
graph TD;
A((P1, V1, T)) -- Isothermal --> B((P2, V2, T));
classDef default fill:#f96,stroke:#333,stroke-width:4px;
style A fill:#1f78b4,stroke:#333,stroke-width:4px;
style B fill:#a6cee3,stroke:#333,stroke-width:4px;
An isothermal process occurs at a constant temperature.
Adiabatic Processes
Adiabatic processes involve no heat exchange with the surroundings. [ P V^\gamma = \text{constant} ]
Isochoric Processes
In an isochoric process, the volume is kept constant. [ \Delta Q = \Delta U ]
Isobaric Processes
In an isobaric process, the pressure is constant. [ W = P \Delta V ]
Cyclic Processes
In cyclic processes, the system returns to its initial state. [ \Delta U = 0 ] [ \Delta Q = \Delta W ]
Thermodynamic Variables and Equations of State
Extensive and Intensive Variables
Extensive variables: Depend on the size of the system (e.g., Volume (V), Mass (m)).
Intensive variables: Independent of the system size (e.g., Pressure (P), Temperature (T)).
Ideal Gas Equation
[ P V = \mu R T ]
Real Gases and their Equations
Real gases deviate from ideal behaviour at high pressures and low temperatures and are described using more complex equations of state.
Specific Heat Capacity
Understanding Specific Heat Capacity
Definition and Formula
Specific heat capacity ((s)) is defined as the amount of heat required to raise the temperature of 1 kg of a substance by 1 K. [ s = \frac{1}{m} \frac{\Delta Q}{\Delta T} ]
Specific Heat at Constant Pressure ((C_p)) and Volume ((C_v))
[ C_p - C_v = R ]
Molar Specific Heat Capacity
Molar specific heat capacity ((C)) is the heat capacity per mole of substance.
Examples and Applications
Specific heat capacities are crucial in understanding climate patterns, cooking, and designing thermal systems.
Reversible and Irreversible Processes
Nature of Processes
Definitions and Differences
Reversible processes can be reversed without leaving any net change in both the system and the surroundings, while irreversible processes cannot.
Quasi-static Processes
A quasi-static process is an idealised process that happens infinitesimally slowly, allowing the system to remain in equilibrium.
Real-World Examples
Examples of irreversible processes include spontaneous heat transfer, friction, and natural gas expansion.
Carnot Cycle and Engines
Carnot Engine
Steps of Carnot Cycle
The Carnot cycle includes two isothermal processes and two adiabatic processes.
Achieving Maximum Efficiency
The Carnot engine is considered the most efficient engine possible between two temperature reservoirs. Its efficiency is given by: [ \eta = 1 - \frac{T_2}{T_1} ]
Importance in Thermodynamics
The Carnot cycle helps set the maximum efficiency for real-world engines and serves as a standard for evaluating other thermodynamic cycles.
Summary and Important Formulas
Key Takeaways
Zeroth Law: Defines temperature.
First Law: Conservation of energy.
Second Law: Entropy and efficiency limits.
Important Formulas
First Law: [ \Delta Q = \Delta U + \Delta W ]
Specific Heat: [ s = \frac{1}{m} \frac{\Delta Q}{\Delta T} ]
Ideal Gas: [ P V = \mu R T ]
Adiabatic Process: [ P V^\gamma = \text{constant} ]
Carnot Efficiency: [ \eta = 1 - \frac{T_2}{T_1} ]
Applications
Understanding these fundamental principles and equations helps in solving thermodynamic problems and designing more efficient systems.
FAQs in Thermodynamics
What is thermodynamics?
The study of heat, temperature, and energy conversion.
What is the Zeroth Law?
It defines temperature as a measure of thermal equilibrium.
What is internal energy?
The total energy stored within a system.
What are isothermal and adiabatic processes?
Isothermal: Constant temperature. Adiabatic: No heat exchange.
This guide provides a concise yet comprehensive overview of Class 11 thermodynamics, ensuring a solid understanding of key concepts and principles.
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