Test Index

Thermodynamics Part 5

Section: Physics

© examsnet.com

Question : 22 of 86

Marks: +1, -0

10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from

P_1

P_2

\alpha

(P_1=21.7\,\text{Pa}.

P_2=30\,\text{Pa}, C_v=21\,\text{J / K . mol}, R=8.3\,\text{J / mol . K})

\alpha

\_\_\_\_

[24 Jan 2026 Shift 2]

21
28
24
15

Solution:

Concept:

Process is isochoric (constant volume).

Formula:

For isochoric process:

W=0

, so

Q=\Delta U = n C_v \Delta T

.

Ideal gas law:

PV=nRT

.

Solution:

From PV diagram, volume is constant at

V=1 \; \text{m}^3

.

Given:

P_1=21.7 \; \text{Pa}

,

P_2=30 \; \text{Pa}

,

C_v=21 \; \text{J/(K . mol)}

,

R=8.3 \; \text{J/(mol . K)}

,

n=10

mol.

Using

PV=nRT

:

n \Delta T = \frac{P_2 V - P_1 V}{R} = \frac{(30-21.7)\times1}{8.3} = \frac{8.3}{8.3}=1

mol·K.

Heat involved:

Q = n C_v \Delta T = C_v \times (n \Delta T) = 21 \times 1 = 21

J.

Thus

\alpha = 21

.

Answer:

Option A: 21

© examsnet.com

Go to Question:

Prev Question

Next Question