Browse by Stream
Discuss Doubts
Home
Search for Exam, Articles, Questions
get app
Go To Your Study Dashboard
  1. Home
  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

Similar Questions

Preparation Products

Knockout JEE Main July 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Rank Booster JEE Main 2020

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 9999/- ₹ 4999/-
Buy Now
Test Series JEE Main July 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 1999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 17999/- ₹ 11999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 19999/-
Buy Now
Exams
Articles
Questions