JEE Main · 2018 · 08 Apr 2018 · Offline · Heat & Thermodynamics

Two moles of an ideal monoatomic gas occupies a volume V at

2 7^{\circ}

C. The gas expands adiabatically to a volume 2V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

Answer: (D)

Solution

For an adiabatic change

T V^{γ - 1}

stays constant, and a monatomic gas has

γ = \frac{5}{3}

. Starting at

T_{1} = 300

K and doubling the volume,

T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} = 300 \times (\frac{1}{2})^{2/3} \approx 189

K. Because the gas expands with no heat exchange its internal energy falls,

Δ U = n C_{V} Δ T = 2 \times \frac{3 R}{2} \times (189 - 300) = 3 R \times (- 111) \approx - 2.7

kJ, so the gas cools to about

189

K with

Δ U \approx - 2.7

kJ.

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option