Question

A rigid diatomic gas \((γ = \frac{7}{5})\) is compressed adiabatically to volume \((\frac{V_i}{32})\), where \(V_i\) is the initial volume. The initial temperature of the gas is '\(T_i\)' K and the final temperature is '\(x\,T_i\)' K. The value of \(x\) is

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(5\)

Hint:

Use T V^(gamma-1) = constant with gamma = 7/5.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In an adiabatic process of an ideal gas, \(TV^{\gamma - 1} = \text{constant}\). Here \(\gamma - 1 = \frac75 - 1 = \frac25\).

Step 2: Apply:
\[ T_iV_i^{2/5} = T_f\left(\frac{V_i}{32}\right)^{2/5} \Rightarrow T_f = T_i\cdot 32^{2/5} \]

Step 3: Evaluate:
\(32 = 2^5\), so \(32^{2/5} = 2^{5\times\frac25} = 2^2 = 4\). Thus \(T_f = 4T_i\), so \(x = 4\).

Step 4: Why the other options are wrong.
\(x = 2\) would need a volume ratio of \(2^{5/2}\), and \(x = 3, 5\) are not powers of 2, so they cannot come from \(32^{2/5}\).

Final Answer:
The value of x is 4, option (C). \[ \boxed{4} \]