Question

A rigid diatomic ideal gas is compressed adiabatically to \((1/32)\) of its initial volume. If the initial temperature of the gas is \(T_{1}\) (in kelvin) and the final temperature is "\(nT_{1}\)", the value of n is \(\{\gamma =1.4\}\)

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

Use T V^(gamma-1) = constant with the given volume ratio and gamma = 1.4.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For a reversible adiabatic process, \(TV^{\gamma-1}\) is constant.

Step 2: Set up
\(T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}\) with \(V_2=\dfrac{V_1}{32}\) and \(\gamma-1=0.4\).

Step 3: Calculation
\(T_2=T_1\left(\dfrac{V_1}{V_2}\right)^{0.4}=T_1(32)^{2/5}\)
Since \(32=2^{5}\), \((32)^{2/5}=2^{2}=4\).

Step 4: Result
\(T_2=4T_1\), so \(n=4\), option (D).

Final Answer:
The value of n is 4. \[ \boxed{\text{(D) }4} \]