# Thermodynamics - Class 11 - Chemistry

Renews every month. Cancel anytime

## Extra Questions - Thermodynamics | NCERT | Chemistry | Class 11

If 2 moles of an ideal monoatomic gas at temperature $T_{0}$ is mixed with 4 moles of another ideal monoatomic gas at temperature $2 \mathrm{~T}_{0}$, then the temperature of the mixture is

A) $\frac{5}{3} T_{0}$

B) $\frac{3}{2} T_{0}$

C) $\frac{4}{3} T_{0}$

D) $\frac{5}{4} T_{0}$

The correct answer is **A) $\frac{5}{3} T_{0}$**. This conclusion is found by considering the **conservation of energy** and the **properties of ideal gases**.

Firstly, since both gases are monoatomic, their degrees of freedom, denoted as $ f $, are similar. For an ideal monoatomic gas, the internal energy ($ U $) is given by: $$ U = \frac{n f R T}{2} $$ where $ n $ is the number of moles, $ R $ is the gas constant, and $ T $ is the temperature. The factor $ \frac{1}{2} $ accounts for the kinetic energy contribution in one degree of freedom (per mole per degree).

When the gases are mixed:

the first gas has 2 moles at temperature $ T_0 $,

the second gas has 4 moles at a temperature of $ 2 T_0 $.

The total internal energy of the mixture, by the conservation of energy, is the sum of the internal energies of the individual gases:
$$
U_{\text{mix}} = U_1 + U_2
$$
We can substitute the values:
$$
\begin{align*}
U_{\text{mix}} &= \frac{2 f R T_0}{2} + \frac{4 f R (2 T_0)}{2} \
&= f R T_0 + 4 f R T_0 \
&= 5 f R T_0
\end{align*}
$$
The internal energy of the mixture can also be written in terms of the unknown final temperature $ T_{\text{mix}} $ of the mixture:
$$
U_{\text{mix}} = \frac{(n_1 + n_2) f R T_{\text{mix}}}{2} \
\Rightarrow \frac{(2 + 4) f R T_{\text{mix}}}{2} = 5 f R T_0
$$
Solving for $ T_{\text{mix}} $:
$$
6 f R T_{\text{mix}} = 10 f R T_0 \
T_{\text{mix}} = \frac{10 T_0}{6} = \frac{5}{3} T_0
$$
Thus, the temperature of the mixture is **$\frac{5}{3} T_0$**.

Two cylindrical rods of lengths $l_{1}$ and $l_{2}$, radii $r_{1}$ and $r_{2}$ have thermal conductivities $k_{1}$ and $k_{2}$ respectively. The ends of the rods are maintained at the same temperature difference. If $l_{1}=2l_{2}$ and $r_{1}=\frac{r_{2}}{2}$, the rates of heat flow in them will be the same if $\frac{k_{1}}{k_{2}}$ is

A. 1

B. 2

C. 4

D. 8

### Improve your grades!

Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.

Sign up now### Improve your grades!

Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.

Sign up now### Improve your grades!

Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.

Sign up now