MHT CET · Physics · Kinetic Theory of Gases
If one mole of an ideal gas \(\left(\gamma=\frac{5}{3}\right)\) is mixed with one mole of a diatomic gas \(\left(\gamma=\frac{7}{5}\right)\). The value of \(\gamma\) for the mixture is
- A \(1.50\)
- B \(1.53\)
- C \(3.07\)
- D \(1.40\)
Answer & Solution
Correct Answer
(A) \(1.50\)
Step-by-step Solution
Detailed explanation
As we know, \(C_v=\frac{3}{5} R T ; C_p=\frac{5 R T}{2}\) for monoatomic gas;
\(C_v=\frac{5}{2} R T ; C_p=\frac{7}{2} R T\) for diatomic gas
Thus, for mixture of 1 mole each,
\(C_v=\frac{\frac{3}{2} R T+\frac{5}{2} R T}{2}\) and \(C_p=\frac{\frac{5}{2} R T+\frac{7}{2} R T}{2}=3 R T\)
Therefore, \(\frac{C_v}{C_p}=\frac{3 R T}{2 R T}=1.5\)
\(C_v=\frac{5}{2} R T ; C_p=\frac{7}{2} R T\) for diatomic gas
Thus, for mixture of 1 mole each,
\(C_v=\frac{\frac{3}{2} R T+\frac{5}{2} R T}{2}\) and \(C_p=\frac{\frac{5}{2} R T+\frac{7}{2} R T}{2}=3 R T\)
Therefore, \(\frac{C_v}{C_p}=\frac{3 R T}{2 R T}=1.5\)
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