JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Heat & Thermodynamics

An ideal monatomic gas of n moles is taken through a cycle

W X Y Z W

consisting of consecutive adiabatic and isobaric quasi-static processes, as shown in the schematic

V - T

diagram. The volume of the gas at

W, X

and

Y

points are, 64 cm³, 125 cm³ and 250 cm³, respectively. If the absolute temperature of the gas

T_{W}

at the point

W

is such that

n R T_{W} = 1

J (R is the universal gas constant), then the amount of heat absorbed (in J) by the gas along the path

X Y

is ____

Answer: 1.6

Solution

Given

n R T_{W} = P_{W} V_{W} = 1

J, so

P_{W} = \frac{1}{64 \times 1 0 ^{- 6}}

Pa.

For adiabatic process WX:

P_{X} = P_{W} (\frac{V _{W}}{V _{X}})^{γ} = P_{W} (\frac{64}{125})^{5/3}

For isobaric process XY (

P_{X} = P_{Y}

Q = n C_{p} Δ T = \frac{5}{2} n R (T_{Y} - T_{X}) = \frac{5}{2} (P_{X} V_{Y} - P_{X} V_{X})

Q = \frac{5}{2} P_{X} (V_{Y} - V_{X}) = \frac{5}{2} \times P_{W} (\frac{64}{125})^{5/3} \times (250 - 125) \times 1 0^{- 6}

Substituting values gives

Q = 1.6

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer