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Question

Chemistry Question on Thermodynamics

55 moles of an ideal gas at 100K100\, K are allowed to undergo reversible compression till its temperature becomes 200K200\, K. If CV=28JK1mol1,caculateΔUC_{V} =28 JK^{-1}mol^{-1} , caculate \Delta\, U and ΔpV\Delta \,pV for this process. (R=8.0JK1mol1R = 8.0 \, JK^{-1}mol^{-1}]

A

ΔU=14kJ:Δ(pV)=4kJ \Delta U = 14\, kJ: \Delta\left(pV\right) = 4\, kJ

B

ΔU=14kJ:Δ(pV)=18kJ \Delta U = 14\, kJ: \Delta\left(pV\right) = 18\, kJ

C

ΔU=2.8kJ:Δ(pV)=0.8kJ \Delta U = 2.8\, kJ: \Delta\left(pV\right) = 0.8\, kJ

D

ΔU=14kJ:Δ(pV)=0.8kJ \Delta U = 14\, kJ: \Delta\left(pV\right) = 0.8\, kJ

Answer

ΔU=14kJ:Δ(pV)=4kJ \Delta U = 14\, kJ: \Delta\left(pV\right) = 4\, kJ

Explanation

Solution

ΔU=nCVm×ΔT\Delta U=nC_{V}m\times\Delta T =5×28×100=14kJ= 5 \times 28 \times 100 = 14\, kJ ΔPV=nRΔT\Delta PV=nR\Delta T =5×8×100=4kJ=5\times 8\times 100=4\, kJ