JEE Advanced · Physics · 13. Thermodynamics
A diatomic ideal gas is compressed adiabatically to \(\frac{1}{32}\) of its initial volume. If the initial temperature of the gas is \(T_i\) \(T_i\) (in kelvin) and the final temperature is \(a T_i\), the value of \(a\) is
- A 2
- B 4
- C 6
- D 8
Answer & Solution
Correct Answer
(B) 4
Step-by-step Solution
Detailed explanation
In adiabatic process,
\(
\begin{aligned}
& T V^{\gamma-1}=\text { constant } \\
& \therefore T_i V_i^{0.4}=T_f V_f^{0.4} \\
& \text { (as } \gamma=1.4 \text { for diatomic gas) } \\
& \text { or } \quad T_i V_i^{0.4}=\left(\alpha T_i\right)\left(\frac{V_i}{32}\right)^{0.4} \\
& \text { or } \quad \alpha(32)^{0.4}=4\end{aligned}
\)
\(\therefore\) The correct answer is 4 .
\(
\begin{aligned}
& T V^{\gamma-1}=\text { constant } \\
& \therefore T_i V_i^{0.4}=T_f V_f^{0.4} \\
& \text { (as } \gamma=1.4 \text { for diatomic gas) } \\
& \text { or } \quad T_i V_i^{0.4}=\left(\alpha T_i\right)\left(\frac{V_i}{32}\right)^{0.4} \\
& \text { or } \quad \alpha(32)^{0.4}=4\end{aligned}
\)
\(\therefore\) The correct answer is 4 .
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