TS EAMCET · Physics · Kinetic Theory of Gases
A gas mixture contains \(n_1\) moles of a monoatomic gas and \(n_2\) moles of gas of rigid diatomic molecules. Each molecule in monoatomic and diatomic gas has 3 and 5 degrees of freedom respectively. If the adiabatic exponent \(\left(\frac{C_p}{C_V}\right)\) for this gas mixture is 1.5 , then the ratio \(\frac{n_1}{n_2}\) will be
- A 1
- B 1.5
- C 2
- D 2.5
Answer & Solution
Correct Answer
(A) 1
Step-by-step Solution
Detailed explanation
For a gas mixture, \[ C_V=\frac{n_1 C_{V_1}+n_2 C_{V_2}}{n_1+n_2} \text { and } C_p=\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1+n_2} \] For monoatomic gas, \[ n=n_1, C_{V_1}=\frac{3}{2} R, \quad C_{p_1}=\frac{5}{2} R \] For diatomic gas,…
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