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Thermodynamics

JEE Main level, single correct, medium

19 of 190

Question 19 of 190

10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from $P_{1}$ to $P_{2}$ is $\alpha$ Joule ( $P_{1}=21.7 \mathrm{~Pa}$ and $\left.P_{2}=30 \mathrm{~Pa}, \mathrm{C}_{v}=21 \mathrm{~J} / \mathrm{K}. \mathrm{mol}, R=8.3 \mathrm{~J} / \mathrm{mol}. \mathrm{K}\right)$ . The value of $\alpha$ is $\_\_\_\_$ .

Show answer and solution

Answer: (A) 21

From the given

P-V

diagram, the process from

P_1

P_2

is an isochoric process because the volume remains constant at

V = 1\text{ m}^3

.
For an isochoric process, the work done

W = 0

.
According to the first law of thermodynamics, the heat involved is

Q = \Delta U + W = \Delta U

.
The change in internal energy is given by

\Delta U = n C_v \Delta T

.
Using the ideal gas equation

PV = nRT

, we have

n\Delta T = \frac{V\Delta P}{R}

.
Substituting this into the expression for

Q

Q = C_v \left( \frac{V(P_2 - P_1)}{R} \right)

Given values:

n = 10

moles,

V = 1\text{ m}^3

P_1 = 21.7\text{ Pa}

P_2 = 30\text{ Pa}

C_v = 21\text{ J/mol}\cdot\text{K}

, and

R = 8.3\text{ J/mol}\cdot\text{K}

Q = 21 \times \frac{1 \times (30 - 21.7)}{8.3}

Q = 21 \times \frac{8.3}{8.3} = 21\text{ J}

.
Thus, the value of

\alpha

is 21.

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