Question

A monatomic gas at 630 K expands adiabatically to 27 times its initial volume. The final temperature of the gas is

• A. \( 30 \) K
• B. \( 130 \) K
• C. \( 170 \) K
• D. \( 70 \) K

Hint:

For an adiabatic process involving an ideal gas, use the relation \( TV^{\gamma - 1} = \text{constant} \). Identify the initial and final states of the gas (temperature and volume). Determine the value of \( \gamma \) for a monatomic gas (\( \gamma = 5/3 \)). Substitute the known values and solve for the unknown final temperature.

Answer 1

The Correct Option is D

For an adiabatic process, the relationship between temperature \( T \) and volume \( V \) is given by \( TV^{\gamma - 1} = \text{constant} \), where \( \gamma \) is the adiabatic index.
For a monatomic gas, \( \gamma = \frac{C_p}{C_v} = \frac{5/2 R}{3/2 R} = \frac{5}{3} \).
Let the initial temperature be \( T_1 = 630 \) K and the initial volume be \( V_1 = V \).
The final volume is \( V_2 = 27 V \).
Let the final temperature be \( T_2 \).
Using the adiabatic relation: \( T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \) \( 630 \cdot V^{(\frac{5}{3} - 1)} = T_2 \cdot (27V)^{(\frac{5}{3} - 1)} \) \( 630 \cdot V^{\frac{2}{3}} = T_2 \cdot (27V)^{\frac{2}{3}} \) \( 630 \cdot V^{\frac{2}{3}} = T_2 \cdot (3^3 V)^{\frac{2}{3}} \) \( 630 \cdot V^{\frac{2}{3}} = T_2 \cdot (3^3)^{\frac{2}{3}} \cdot V^{\frac{2}{3}} \) \( 630 \cdot V^{\frac{2}{3}} = T_2 \cdot 3^{3 \times \frac{2}{3}} \cdot V^{\frac{2}{3}} \) \( 630 \cdot V^{\frac{2}{3}} = T_2 \cdot 3^2 \cdot V^{\frac{2}{3}} \) \( 630 = T_2 \cdot 9 \) \( T_2 = \frac{630}{9} = 70 \) K The final temperature of the gas is 70 K.