Question

A gas is suddenly compressed such that its absolute temperature is doubled. If the ratio of the specific heat capacities of the gas is 1.5, then the percentage decrease in the volume of the gas is

• A. 30
• B. 50
• C. 25
• D. 75

Hint:

For adiabatic processes: $TV^{\gamma-1} = C$, $PV^\gamma = C$, and $T^\gamma P^{1-\gamma} = C$. Ensure $\gamma$ is correctly identified ($C_p/C_v$).

Answer 1

The Correct Option is D

Step 1: Identify Process: "Suddenly compressed" implies an adiabatic process. The relationship between Temperature ($T$) and Volume ($V$) in an adiabatic process is given by: \[ T V^{\gamma - 1} = \text{Constant} \] where $\gamma = 1.5$.
Step 2: Set up Equation: Let initial state be $(T_1, V_1)$ and final state be $(T_2, V_2)$. Given $T_2 = 2T_1$. \[ T_1 V_1^{1.5 - 1} = T_2 V_2^{1.5 - 1} \] \[ T_1 V_1^{0.5} = (2T_1) V_2^{0.5} \] \[ V_1^{0.5} = 2 V_2^{0.5} \]
Step 3: Solve for Volume Ratio: Squaring both sides: \[ (V_1^{0.5})^2 = (2 V_2^{0.5})^2 \] \[ V_1 = 4 V_2 \implies V_2 = \frac{V_1}{4} = 0.25 V_1 \]
Step 4: Calculation Process Percentage Decrease: Decrease in volume $= V_1 - V_2 = V_1 - 0.25V_1 = 0.75V_1$. Percentage decrease $= \frac{0.75 V_1}{V_1} \times 100 = 75%$.