Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

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)

best_answer

\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

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Mains 2026 LIVE: JEE Advanced registration closes on May 2; exam date and timing, BTech cutoffs
anam-khan

JEE Main 2026 toppers reveal success mantras; practice, precision, and perfect strategy
anam-khan

JEE Main 2026 Toppers List: 26 in AIR 1 with 100 percentile, most from Andhra Pradesh, Telangana

Latest Question