Question

In a thermodynamic process, helium gas obeys the law $TP^{-\frac{2}{5}} =$ constant. If the temperature of 2 moles of the gas is raised from $T$ to $3T, then:

  • Heat given to the gas is $9RT$
  • Heat given to the gas is zero
  • Increase in internal energy is $6RT$
  • Work done by the gas is $-6RT$

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Answer

The correct options are:

  • B: Heat given to the gas is zero
  • C: Increase in internal energy is $6RT$
  • D: Work done by the gas is $-6RT$

Given the law: $$ TP^{-\frac{2}{5}} = \text{constant} $$

This can be rewritten as: $$ (PV) \cdot P^{-\frac{2}{5}} = \text{constant} \quad \text{or} \quad PV^{\frac{5}{3}} = \text{constant} $$

When we compare this with the general isothermal relationship: $$ PV^x = \text{constant} $$

We find that: $$ x = \frac{5}{3} $$

The molar heat capacity for this process is given by: $$ C = C_v + \frac{R}{1 - x} $$

Substituting $ x = \frac{5}{3} $: $$ C = \frac{3}{2} R + \frac{R}{1 - \frac{5}{3}} = 0 $$

Hence, the heat given to the gas is zero: $$ Q = n C \Delta T = 0 $$

Next, the relationship between internal energy change (ΔU) and work done (W) is: $$ \Delta U = -W $$

Therefore: $$ \Delta U = n C_v \Delta T $$

Substituting the values: $$ \Delta U = 2 \left( \frac{3}{2} R \right) (3T - T) = 6RT $$

So, the increase in internal energy is $6RT$ and the work done by the gas is $-6RT$.


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