In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation $\mathrm{VT}=\mathrm{K}$, where K is a constant. In this process the temperature of the gas is increased by $\Delta T$. The amount of heat absorbed by gas is ($R$ is gas constant)

(A) $\frac{1}{2} \mathrm{R} \Delta \mathrm{T}$

(B) $\frac{1}{2} \mathrm{KR} \Delta \mathrm{T}$

(D) $\frac{2 \mathrm{~K}}{3} \Delta \mathrm{T}$

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Heat & Thermodynamics

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation

VT = K

, where K is a constant. In this process the temperature of the gas is increased by

Δ T

. The amount of heat absorbed by gas is (

R

is gas constant)

Answer: (A)

Since

V T = K

and, for one mole,

P V = R T

, eliminating

T

gives

V \cdot \frac{P V}{R} = K

, so

P V^{2}

is constant — a polytropic process with exponent

x = 2

. Its molar heat capacity is

C = C_{v} + \frac{R}{1 - x} = \frac{3 R}{2} + \frac{R}{1 - 2} = \frac{3 R}{2} - R = \frac{R}{2}

. The heat absorbed for a rise

Δ T

of one mole is

Δ Q = C Δ T = \frac{1}{2} R Δ T

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option