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  2. Answer please! - Chemical Thermodynamics - JEE Main
# Engineering

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

  • Option 1)

    \Delta U=14 kJ; \Delta (pV)=18kJ

  • Option 2)

    \Delta U=14J; \Delta (pV)=0.8J

  • Option 3)

    \Delta U=14 kJ; \Delta (pV)=4kJ

  • Option 4)

    \Delta U=2.8 kJ; \Delta (pV)=0.8kJ

Answers (1)
S solutionqc

\Delta U=nCv,m \;\; \Delta T

=5 \times 28\times (200-100)

=14000J=14KJ

\Delta(pV)=P_{2}V_{2}-P_{1}V_{1}=nRT_{2}-nRT_{1}

                nR(T_{2}-T_{1})=5\times 8 \times 100=4000J=4kJ

ans is  \Delta U=14 kJ; \Delta (pV)=4kJ


Option 1)

\Delta U=14 kJ; \Delta (pV)=18kJ

Option 2)

\Delta U=14J; \Delta (pV)=0.8J

Option 3)

\Delta U=14 kJ; \Delta (pV)=4kJ

Option 4)

\Delta U=2.8 kJ; \Delta (pV)=0.8kJ

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