Question

A certain volume of a gas at 300 K expands adiabatically until its volume is doubled. The resultant fall in temperature of the gas is nearly (The ratio of the specific heats of the gas is 1.5)

• A. \( 88 \, {K} \)
• B. \( 77 \, {K} \)
• C. \( 67 \, {K} \)
• D. \( 54 \, {K} \)

Hint:

Remember that in adiabatic processes, no heat is exchanged with the surroundings, so all changes in temperature are due to changes in volume or pressure.

Answer 1

The Correct Option is A

For an ideal gas undergoing an adiabatic process, the relationship between temperature and volume is given by \( T V^{\gamma-1} = {constant} \), where \( \gamma \) is the ratio of specific heats. 
Given \( \gamma = 1.5 \) and initial temperature \( T_i = 300 \, {K} \), the final temperature \( T_f \) when the volume is doubled can be found by setting: \[ T_i V^{\gamma-1} = T_f (2V)^{\gamma-1} \] Solving for \( T_f \): \[ 300 \times 1 = T_f \times 2^{0.5} \Rightarrow T_f \approx 212 \, {K} \] The temperature drop \( \Delta T \) is: \[ \Delta T = 300 \, {K} - 212 \, {K} = 88 \, {K} \]