Q171 / 180 · COMEDK 2023 (Evening Shift) · Physics
If the ratio of specific heat of a gas at constant pressure to that at constant volume is $\gamma$, the change in internal energy of a mass of a gas when the volume changes from $\mathrm{V}$ to $3 \mathrm{~V}$ at constant pressure is
- A$\dfrac{2 P V}{\gamma}$
- B$\dfrac{R}{(\gamma-1)}$
- Correct answerC$\dfrac{2 P V}{(\gamma-1)}$
- D$2 P V$
Solution
The change in internal energy $\Delta U$ for an ideal gas is given by the formula $\Delta U = n C_{v} \Delta T$.
For an ideal gas, $C_{v} = \dfrac{R}{\gamma - 1}$. Thus, $\Delta U = n \left( \dfrac{R}{\gamma - 1} \right) \Delta T$.
Using the ideal gas law $PV = nRT$, at constant pressure $P$, we have $P \Delta V = nR \Delta T$.
Substituting $nR \Delta T = P \Delta V$ into the expression for internal energy, we get $\Delta U = \dfrac{P \Delta V}{\gamma - 1}$.
Given the volume changes from $V$ to $3V$, the change in volume is $\Delta V = 3V - V = 2V$.
Therefore, $\Delta U = \dfrac{P(2V)}{\gamma - 1} = \dfrac{2PV}{\gamma - 1}$.
Answer: $\dfrac{2 P V}{(\gamma-1)}$
For an ideal gas, $C_{v} = \dfrac{R}{\gamma - 1}$. Thus, $\Delta U = n \left( \dfrac{R}{\gamma - 1} \right) \Delta T$.
Using the ideal gas law $PV = nRT$, at constant pressure $P$, we have $P \Delta V = nR \Delta T$.
Substituting $nR \Delta T = P \Delta V$ into the expression for internal energy, we get $\Delta U = \dfrac{P \Delta V}{\gamma - 1}$.
Given the volume changes from $V$ to $3V$, the change in volume is $\Delta V = 3V - V = 2V$.
Therefore, $\Delta U = \dfrac{P(2V)}{\gamma - 1} = \dfrac{2PV}{\gamma - 1}$.
Answer: $\dfrac{2 P V}{(\gamma-1)}$
Hence, the correct answer is option: 'C'.