JEE Mains · Physics · STD 11 - 11. thermodynamics
A sample of gas at temperature \(T\) is adiabatically expanded to double its volume. The work done by the gas in the process is \(\left(\right.\) given, \(\left.\gamma=\frac{3}{2}\right)\) :
- A \(W=T R[\sqrt{2}-2]\)
- B \(W=\frac{T}{R}[\sqrt{2}-2]\)
- C \(W=\frac{R}{T}[2-\sqrt{2}]\)
- D \(W=R T[2-\sqrt{2}]\)
Answer & Solution
Correct Answer
(D) \(W=R T[2-\sqrt{2}]\)
Step-by-step Solution
Detailed explanation
\(T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}\) \(T V^{1 / 2}=T_2(2 V)^{1 / 2}\) \(T_2=\frac{T}{\sqrt{2}}\) \(W=\frac{R\left(T_1-T_2\right)}{\gamma-1}=\frac{R\left(T-\frac{T}{\sqrt{2}}\right)}{\frac{1}{2}}=R T(2-\sqrt{2})\)
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