KTG & Thermodynamics_JEE Advanced_2024_S1 - Competishun – Live JEE & NEET Courses and Exam Prep

JEE Advanced 2024

Paper-1 2024

Question

One mole of a monatomic ideal gas undergoes the cyclic process $J \rightarrow K \rightarrow L \rightarrow M \rightarrow J$, as shown in the $P-T$ diagram.

Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [R is the gas constant.]

Select the correct option:

P → 1; Q → 3; R → 5; S → 4

P → 4; Q → 3; R → 5; S → 2

P → 4; Q → 1; R → 2; S → 2

P → 2; Q → 5; R → 3; S → 4

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

From $M \rightarrow J$, isothermal
$ \begin{aligned} & \mathrm{W}=\mathrm{nRT} \ln 2 \\ & =\mathrm{RT}_0 \ln 2 \\ & \Delta \mathrm{U}=0 \end{aligned} $
From $\mathrm{J} \rightarrow \mathrm{K}$, isobaric
$ \begin{aligned} & W=n R\left(3 T_0-T_0\right) \\ & W=2 R T_0 \\ & \Delta U=n C_v \Delta T \\ & =\frac{3 R}{2} \times 2 T_0=3 R T_0 \end{aligned} $
From $K \rightarrow L$, isothermal
$ \begin{aligned} & W=-n R\left(3 T_0\right) \ln 2 \\ & W=-3 R T_0 \ln 2 \\ & \Delta U=0 \\ & Q=-3 R T_0 \ln 2 \end{aligned} $
From $L \rightarrow M$, isobaric
$ \begin{aligned} & W=n R\left(T_0-3 T_0\right) \\ & W=-2 R T_0 \\ & \Delta U=n C_v \Delta T \\ & =n \times \frac{3 R}{2} \times\left(-2 T_0\right)=-3 R T_0 \end{aligned} $
For $P$ :-
$ \begin{aligned} & W_{\text {net }}=R T_0 \ln 2+2 R T_0-3 R T_0 \ln 2-2 R T_0 \\ & =-2 R T_0 \ln 2 \\ & P \rightarrow 4 \\ & \text { for } Q \rightarrow 3 \\ & \text { for } R \rightarrow 5 \\ & \text { for } S \rightarrow 2 \end{aligned} $

Question Tags

JEE Advance

Physics

Easy

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