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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main · Physics · Kinetic Theory of Gases

A gaseous mixture consists of 16 ~g of helium and 16 ~g of oxygen....

A gaseous mixture consists of 16 g16 \mathrm{~g} of helium and 16 g16 \mathrm{~g} of oxygen. The ratio Cp/CVC_{p} / C_{V} of the mixture is

Options

  1. A1.41.4
  2. B1.541.54
  3. C1.591.59
  4. D1.621.62

Correct answer

D. 1.621.62

Step-by-step solution

Mass of helium, m1=16 gm_{1}=16 \mathrm{~g} Mass of oxygen, m2=16 g∴m_{2}=16 \mathrm{~g}\therefore Number of moles of He gas, n_{1}=m14=164=\frac{m_{1}}{4}=\frac{16}{4}=4 Number of moles of O2gas\mathrm{O}_{2} \mathrm{gas}, n_{2}=m232=1632=12\frac{m_{2}}{32}=\frac{16}{32}=\frac{1}{2}We know that for the mixture, CV=n1CV+n2CV2n1+n2, where CV=fR2 and Cp=n1Cp1+n2Cp2n1+n2, where Cp=(f2+1)R\begin{aligned} C_{V} &=\frac{n_{1} C_{V}+n_{2} C_{V_{2}}}{n_{1}+n_{2}}, \text { where } C_{V}=\frac{f R}{2} \\ \text { and } C_{p} &=\frac{n_{1} C_{p_{1}}+n_{2} C_{p_{2}}}{n_{1}+n_{2}}, \text { where } C_{p}=\left(\frac{f}{2}+1\right) R \end{aligned} For He gas, f=3,n1=4f=3, n_{1}=4 For O2\mathrm{O}_{2} gas, f=5,n2=12f=5, n_{2}=\frac{1}{2}∴CpCV=n1Cp1+n2Cp2n1CV1+n2CV2=4×(32+1)R+12(52+1)R4×32R+12×52R=(4×52R)+(12×72R)(4×32R)+(12×52R)=4729=1.62\begin{aligned} \therefore \quad\frac{C_{p}}{C_{V}}&=\frac{n_{1} C_{p_{1}}+n_{2} C_{p_{2}}}{n_{1} C_{V_{1}}+n_{2} C_{V_{2}}}\\ &=\frac{4 \times\left(\frac{3}{2}+1\right) R+\frac{1}{2}\left(\frac{5}{2}+1\right) R}{4 \times \frac{3}{2} R+\frac{1}{2} \times \frac{5}{2} R}\\ &=\frac{\left(4 \times \frac{5}{2} R\right)+\left(\frac{1}{2} \times \frac{7}{2} R\right)}{\left(4 \times \frac{3}{2} R\right)+\left(\frac{1}{2} \times \frac{5}{2} R\right)}=\frac{47}{29}\\ &=1.62 \end{aligned}$
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