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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Physics
Chapter
Kinetic Theory of Gases

BITSAT 2022 · Physics · Kinetic Theory of Gases

One mole of an ideal diatomic gas undergoes a process as shown in...

One mole of an ideal diatomic gas undergoes a process as shown in the figure. The molar specific heat of the gas in the process is One mole of an ideal diatomic gas undergoes a process as shown in the figure. The molar specific heat of the gas in the

Options

  1. A3R/23 R / 2
  2. BR/2R / 2
  3. C5R/25 R / 2
  4. D7R/27 R / 2

Correct answer

A. 3R/23 R / 2

Step-by-step solution

In the curve, V∝1TV \propto \frac{1}{T}⇒V=m1T where, m= slope of  curve.  and ⇒VT= constant  and ∴TdV+VdT=0 and ⇒dV=−VTdT As, Q=ΔU+WCdT=CVdT+pdVCdT=CVdT−pTVdT⇒C=CV−pVT and =CV−R(∵pVT=R)\begin{aligned}& \Rightarrow V=m \frac{1}{T} \text{ where, } m=\text{ slope\text{ of } curve. }\\ \text{ and } \Rightarrow V T=\text{ constant }\\ \text{ and } \therefore \quad T d V+V d T=0 \\ \text{ and } \Rightarrow \quad d V=-\frac{V}{T} d T \\ \text{ As, } Q=\Delta U+W \\ & C d T=C_V d T+p d V \\ & C d T=C_V d T-\frac{p}{T} V d T \\ & \Rightarrow \quad C=C_V-\frac{p V}{T}\\ \text{ and } =C_V-R \quad\left(\because \frac{p V}{T}= R\right) \\ & \end{aligned}Fordiatomicgas,For diatomic gas, C_V=52\frac{5}{2} R \therefore \quad C=C_V-R=52R−R=3R2\frac{5}{2} R-R= \frac{3 R}{2}
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