JEE Advanced 2025 Paper 2 · Q11 · Kinetic Theory of Gases

Question

An ideal monatomic gas of $n$ moles is taken through a cycle WXYZW consisting of consecutive adiabatic and isobaric quasi-static processes (V-T diagram with W at low V/T, X reached by adiabatic from W, Y reached by isobaric expansion from X, Z by adiabatic from Y, and back to W by isobaric compression). The volumes at W, X and Y are $64$ cm³, $125$ cm³ and $250$ cm³ respectively. If the absolute temperature at 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 XY is __________.

SolveKar AI · Accepted Answer

WX is adiabatic for monatomic gas ($\gamma = 5/3$): $P_W V_W^\gamma = P_X V_X^\gamma$. $P_W = n R T_W/V_W = 1	ext{ J}/(64 	imes 10^{-6}	ext{ m}^3) = (1/64) 	imes 10^6$ Pa. $P_X = P_W (V_W/V_X)^{5/3} = P_W (64/125)^{5/3}$. $(64/125)^{1/3} = 4/5$; raised to 5: $(4/5)^5 = 1024/3125$. XY is isobaric expansion ($P_Y = P_X$), volume goes from $V_X$ to $V_Y$. Heat absorbed $Q = n C_p \Delta T = (5/2)\,n R \Delta T = (5/2)\,P_X (V_Y - V_X)$ (using ideal gas $n R \Delta T = P\,\Delta V$ at constant $P$). $Q = (5/2)((1/64) 	imes 10^6)(1024/3125)(250 - 125) 	imes 10^{-6} = (5/2)(125/64)(1024/3125) = (5 \cdot 125 \cdot 1024)/(2 \cdot 64 \cdot 3125) = 640000/400000 = 1.6$ J.

JEE Advanced 2025 Paper 2 · Q11 · Kinetic Theory of Gases

Solution