Q162 / 180 · COMEDK 2025 (10 May Shift 2) · Physics
An ideal gas is expanding such that $P T^2=$ constant. The coefficient of volume expansion of the gas is
- A$\dfrac{T}{3}$
- B$3 T$
- C$\dfrac{2}{T}$
- Correct answerD$\dfrac{3}{T}$
Solution
The equation of state for an ideal gas is $PV = nRT$.
Given $PT^{2} = C$, where $C$ is a constant, we can write $P = C T^{-2}$.
Substituting this into the ideal gas equation, we get $(C T^{-2}) V = nRT$.
Rearranging for $V$, we have $V = \dfrac{nR}{C} T^{3}$.
The coefficient of volume expansion $\gamma$ is defined as $\gamma = \dfrac{1}{V} \left( \dfrac{dV}{dT} \right)_{P}$.
Differentiating $V$ with respect to $T$ from the relation $V = \dfrac{nR}{C} T^{3}$, we get $\dfrac{dV}{dT} = 3 \dfrac{nR}{C} T^{2}$.
Substituting $V$ and $\dfrac{dV}{dT}$ into the expression for $\gamma$:
$\gamma = \dfrac{1}{\left( \dfrac{nR}{C} T^{3} \right)} \left( 3 \dfrac{nR}{C} T^{2} \right) = \dfrac{3}{T}$.
Answer: $\dfrac{3}{T}$
Given $PT^{2} = C$, where $C$ is a constant, we can write $P = C T^{-2}$.
Substituting this into the ideal gas equation, we get $(C T^{-2}) V = nRT$.
Rearranging for $V$, we have $V = \dfrac{nR}{C} T^{3}$.
The coefficient of volume expansion $\gamma$ is defined as $\gamma = \dfrac{1}{V} \left( \dfrac{dV}{dT} \right)_{P}$.
Differentiating $V$ with respect to $T$ from the relation $V = \dfrac{nR}{C} T^{3}$, we get $\dfrac{dV}{dT} = 3 \dfrac{nR}{C} T^{2}$.
Substituting $V$ and $\dfrac{dV}{dT}$ into the expression for $\gamma$:
$\gamma = \dfrac{1}{\left( \dfrac{nR}{C} T^{3} \right)} \left( 3 \dfrac{nR}{C} T^{2} \right) = \dfrac{3}{T}$.
Answer: $\dfrac{3}{T}$
Hence, the correct answer is option: 'D'.