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Thermodynamics Part 5

Section: Physics

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Question : 32 of 86

Marks: +1, -0

The volume of an ideal gas increases 8 times and temperature becomes

\left(\frac{1}{4}\right)^{\;\text{th }\;}

of initial temperature during a reversible change. If there is no exchange of heat in this process (

\Delta Q=0

) then identify the gas from the following options (Assuming the gases given in the options are ideal gases):

[22 Jan 2026 Shift 1]

$\mathrm{NH}_3$
$\mathrm{O}_2$
$\mathrm{CO}_2$
He

Solution:

Concept:

For an adiabatic process (

\Delta Q=0

), the relation between temperature and volume for an ideal gas is

T V^{\gamma-1}=

constant, where

\gamma = C_p/C_v

.

Explanation:

Given:

V_2 = 8V_1

and

T_2 = T_1/4

.

Using

T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}

:

T_1 V_1^{\gamma-1} = (T_1/4)(8V_1)^{\gamma-1}

.

Cancel

T_1

and

V_1^{\gamma-1}

:

1 = \frac{1}{4} \cdot 8^{\gamma-1}

.

\Rightarrow 4 = 8^{\gamma-1} \Rightarrow 2^2 = (2^3)^{\gamma-1} \Rightarrow 2^2 = 2^{3(\gamma-1)}

.

Equate exponents:

2 = 3(\gamma-1) \Rightarrow \gamma-1 = \frac{2}{3} \Rightarrow \gamma = \frac{5}{3}

.

\gamma = \frac{5}{3}

corresponds to a monatomic gas.

Among options, only He is monatomic.

Answer:

D. He

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