Question

An ideal gas expands adiabatically, ($\gamma = 1.5$). To reduce the r.m.s. velocity of the molecules 4 times, the gas has to be expanded ______.

• A. 256 times
• B. 128 times
• C. 64 times
• D. 8 times

Hint:

Adiabatic expansions always cool a gas down because the gas expends internal energy to do the work of expansion. A massive expansion ($256\times$) is required to achieve a small velocity reduction ($4\times$) because velocity scales with the square root of temperature!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We must link the macroscopic thermodynamic adiabatic expansion of a gas to the microscopic kinetic theory of r.m.s. velocity to find the required volume change.

Step 2: Detailed Explanation:
1. Relate r.m.s. velocity to Temperature:
The r.m.s. velocity of an ideal gas is $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$.
This means velocity is proportional to the square root of the absolute temperature: $v_{\text{rms}} \propto \sqrt{T}$.
The problem states we want to reduce the velocity 4 times:
$v_2 = \frac{v_1}{4}$
Since $v \propto \sqrt{T}$, to reduce the velocity by a factor of 4, the temperature must be reduced by a factor of $4^2 = 16$.
$T_2 = \frac{T_1}{16}$
So, the final temperature ratio is $\frac{T_1}{T_2} = 16$.
2. Relate Temperature to Volume via Adiabatic Expansion:
For an adiabatic process, the relationship between Temperature and Volume is:
$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$
Rearrange to isolate the volume ratio:
$\left(\frac{V_2}{V_1}\right)^{\gamma - 1} = \frac{T_1}{T_2}$
We are given the adiabatic index $\gamma = 1.5$.
Therefore, the exponent is $\gamma - 1 = 1.5 - 1 = 0.5 = \frac{1}{2}$.
Substitute the known values into the equation:
$\left(\frac{V_2}{V_1}\right)^{1/2} = 16$
To solve for the volume expansion ratio, square both sides:
$\frac{V_2}{V_1} = 16^2$
$\frac{V_2}{V_1} = 256$
The volume must be expanded 256 times its original size.

Step 3: Final Answer:
The gas has to be expanded 256 times, matching option (a).