5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If $\mathrm{C}_{\mathrm{v}}=28 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, calculate $\Delta \mathrm{U}$ and $\Delta \mathrm{pV}$ for this process. ($\mathrm{R}=8.0 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$)

(B) $\Delta \mathrm{U}=14 \mathrm{~kJ} ; \Delta(\mathrm{pV})=4 \mathrm{~kJ}$

(A) $\Delta \mathrm{U}=2.8 \mathrm{~kJ} ; \Delta(\mathrm{pV})=0.8 \mathrm{~kJ}$

(C) $\Delta \mathrm{U}=14 \mathrm{~kJ} ; \Delta(\mathrm{pV})=18 \mathrm{~kJ}$

(D) $\Delta \mathrm{U}=14 \mathrm{~kJ} ; \Delta(\mathrm{pV})=0.8 \mathrm{~J}$

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Chemical Thermodynamics

5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If

C_{v} = 28 JK^{- 1} mol^{- 1}

, calculate

Δ U

and

Δ pV

for this process. (

R = 8.0 J K^{- 1} mol^{- 1}

)

Answer: (B)

For an ideal gas the internal energy depends only on temperature, so

Δ U = n C_{v} Δ T = 5 \times 28 \times (200 - 100) = 14000 J = 14 kJ

. For the

p V

product, the gas law

p V = n R T

makes

Δ (p V) = n R Δ T = 5 \times 8.0 \times 100 = 4000 J = 4 kJ

. So the process has

Δ U = 14 kJ

and

Δ (p V) = 4 kJ

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option