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

JEE Main · Physics · Kinetic Theory of Gases

One mole of a monoatomic ideal gas undergoes a quasistatic process,...

One mole of a monoatomic ideal gas undergoes a quasistatic process, which is depicted by a straight line joining points (V0,T0)\left(V_{0}, T_{0}\right) and (2V0,3T0)\left(2 V_{0}, 3 T_{0}\right) in a V⋅T{V} \cdot T diagram. What is the value of the heat capacity of the gas at the point \left(V_{0}, T_{0}\right) ?

Options

  1. ARR
  2. B32R\frac{3}{2} R
  3. C2R2 R
  4. D0

Correct answer

C. 2R2 R

Step-by-step solution

Heat capacity of an ideal gas in a thermodynamic process, ∵\because Ideal gas is monoatomic,CV=fR2=3R2Cprocess =CV+pn⋅dVdT=32R+pn⋅V02T∴Cprocess =32R+nRTn2T[∵pV=nRT]=32R+R2=2R\begin{aligned}C_{V} &=\frac{f R}{2}=\frac{3 R}{2}\\ C_{\text {process }} &=C_{V}+\frac{p}{n} \cdot \frac{d V}{d T}\\ &=\frac{3}{2}R+\frac{p}{n} \cdot \frac{V_{0}}{2 T}\\ \therefore C_{\text {process }}=\frac{3}{2}R+\frac{n R T}{n 2 T}&[\because p V=n R T] \\ =& \frac{3}{2}R+\frac{R}{2}=2 R \end{aligned}
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