Q54 / 270 · NDA 2026 (Phase 1) · General Ability
In a certain process, the relation between pressure $(P)$ and volume $(V)$, for an ideal gas, is given by $PV^2 = \text{constant}$. If the initial temperature is $T_1$ and the final temperature is $T_2$; and the initial volume is $V_1$ and the final volume is $V_2$, then which one of the following is correct?
- A$\dfrac{T_1}{T_2} = \dfrac{V_1}{V_2}$
- B$\dfrac{T_1}{T_2} = \left(\dfrac{V_1}{V_2}\right)^2$
- Correct answerC$\dfrac{T_1}{T_2} = \dfrac{V_2}{V_1}$
- D$\dfrac{T_1}{T_2} = \left(\dfrac{V_2}{V_1}\right)^2$
Solution
The given process equation is $PV^2 = \text{constant}$.
For an ideal gas, the equation of state is $PV = nRT$, which gives $P = \dfrac{nRT}{V}$.
Substituting the expression for $P$ into the process equation:
$\left(\dfrac{nRT}{V}\right) V^2 = \text{constant}$
$T V = \dfrac{\text{constant}}{nR} = \text{constant}$
This implies that $T_1 V_1 = T_2 V_2$.
Rearranging the terms, we get:
$\dfrac{T_1}{T_2} = \dfrac{V_2}{V_1}$
Answer: $\dfrac{T_1}{T_2} = \dfrac{V_2}{V_1}$
For an ideal gas, the equation of state is $PV = nRT$, which gives $P = \dfrac{nRT}{V}$.
Substituting the expression for $P$ into the process equation:
$\left(\dfrac{nRT}{V}\right) V^2 = \text{constant}$
$T V = \dfrac{\text{constant}}{nR} = \text{constant}$
This implies that $T_1 V_1 = T_2 V_2$.
Rearranging the terms, we get:
$\dfrac{T_1}{T_2} = \dfrac{V_2}{V_1}$
Answer: $\dfrac{T_1}{T_2} = \dfrac{V_2}{V_1}$
Hence, the correct answer is option: 'C'.