JEE Mains · Physics · STD 11 - Kinetic theory of gases
- JEE Mains 2020
- Hard
- MCQ
Two moles of an ideal gas with \(\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{5}{3}\) are mixed with \(3\) moles of another ideal gas with \(\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{4}{3} .\) The value of \(\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}\) for the mixture is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(1.42\)
Step-by-step Solution
Detailed explanation
For \(1^{\text {st }}\) gas \(\frac{C_{P_{1}}}{C_{V_{1}}}=\frac{5}{3} \Rightarrow C_{P_{1}}=5 x\) and \(C_{V_{1}}=3 x\)
For \(2^{\text {nd }}\) gas \(\frac{C_{P_{2}}}{C_{V_{2}}}=\frac{4}{3} \Rightarrow C_{P_{2}}=4 x\) and \(C_{V_{2}}=3 x\)
Now for mixture \(\mathrm{C}_{\mathrm{P}}=\frac{\mathrm{n}_{1} \mathrm{C}_{\mathrm{P}}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{B}_{2}}}{\mathrm{n}_{1}+\mathrm{n}_{2}}=\frac{17 \mathrm{R}}{5}\)
\(C_{V}=\frac{n_{1} C_{V_{1}}+n_{2} C_{V_{3}}}{n_{1}+n_{2}}=\frac{12 R}{5}\)
\(\Rightarrow \frac{C_{p}}{C_{v}}=\frac{2(5 x)+3(4 x)}{2(3 x)+3(3 x)}=\frac{17}{12}\)
\(\Rightarrow \frac{C_{P}}{C_{V}} \approx 1.42\)
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