Verified Solution • 2025–26 Session moderate

QPages 77-80: Q9: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q9 Solution: An ideal gas is taken through a process in which the pressure…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often differentiate $pV=kV^2$ as $d(pV)=kV,dV$ (treating $dp$ as zero) and miss the extra $kV,dV$ from $Vdp$, which actually gives $d(pV)=2kV,dV$. Remember that $dp=k,dV$, so $p,dV+V,dp=2kV,dV$, yielding the correct $ rac{R}{2}dT$ work term.

An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation p=kVp = kV. Show that the molar heat capacity of the gas for the process is given by C=Cv+R2C = C_v + \frac{R}{2}.

Step-by-Step Verified Solution

**Answer:** C=Cv+R2C = C_v + \frac{R}{2}. **Step-by-Step Explanation:** Given: P=kVP = kV. Using PV=nRTPV = nRT: kV2=nRTkV^2 = nRT. For n=1n=1: RdT=2kVdVRdT = 2kVdV. Work done: dW=P dV=kV dV=R2dTdW = P\,dV = kV\,dV = \frac{R}{2}dT. ΔU=CvdT\Delta U = C_v dT. dQ=ΔU+dW=CvdT+R2dT=(Cv+R2)dTdQ = \Delta U + dW = C_v dT + \frac{R}{2}dT = \left(C_v + \frac{R}{2}\right)dT. So C=Cv+R2C = C_v + \frac{R}{2} [citation:12].

All Questions in this Chapter

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