Thermodynamics

JEE Main level, single correct, medium

87 of 190

Question 87 of 190

Two different adiabatic paths for the same gas intersect two isothermal curves as shown in P-V diagram. The relation between the ratio VaVd\frac{V_a}{V_d} and the ratio VbVc\frac{V_b}{V_c} is:

Show answer and solution

Answer: (B) VaVd=VbVc\frac{V_a}{V_d}=\frac{V_b}{V_c}

For adiabatic process
TVγ−1=\mathrm{TV}^{\gamma-1}= constant
Ta⋅Vaγ−1=Td⋅Vdγ−1(VaVd)γ−1=TdTaTb⋅Vbγ−1=Tc⋅Vcγ−1(VbVc)γ−1=TcTbVaVd=VbVc(∵Td=TcTa=Tb)\begin{aligned} & \mathrm{T}_{\mathrm{a}} \cdot \mathrm{V}_{\mathrm{a}}^{\gamma-1}=\mathrm{T}_{\mathrm{d}} \cdot \mathrm{V}_{\mathrm{d}}^{\gamma-1} \\ & \left(\frac{\mathrm{V}_{\mathrm{a}}}{\mathrm{V}_{\mathrm{d}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{d}}}{\mathrm{T}_{\mathrm{a}}} \\ & \mathrm{T}_{\mathrm{b}} \cdot \mathrm{V}_{\mathrm{b}}^{\gamma-1}=\mathrm{T}_{\mathrm{c}} \cdot \mathrm{V}_{\mathrm{c}}^{\gamma-1} \\ & \left(\frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{c}}}{\mathrm{T}_{\mathrm{b}}} \\ & \frac{\mathrm{V}_{\mathrm{a}}}{\mathrm{V}_{\mathrm{d}}}=\frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}} \quad\left(\begin{array}{c}\because \mathrm{T}_{\mathrm{d}}=\mathrm{T}_{\mathrm{c}} \\ \mathrm{T}_{\mathrm{a}}=\mathrm{T}_{\mathrm{b}}\end{array}\right)\end{aligned}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Thermodynamics (Thermodynamic Processes), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.