Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
1 Mole of CO2 gas at 300 K expanded under the reversible adiabatic condition such that its volume becomes 27 times. The magnitude of work done inkJ/mol is:
(Given γ=1.33 and  Cv=25.10Jmol1K1forCO2 )
report your answer by rounding it up to nearest whole number
ChemistryThermodynamics (C)JEE Main
Solution:
1935 Upvotes Verified Answer
The correct answer is: 5
Number of moles =1

T1=300 K

V2=27V1

TV1γ-1=T2V2γ-1

T1T2=V2V1r1

T2=30012713

T2=100 K

Adiabatic condition, q=0;ΔE=w=nCvT2T1

w=1×25.10100300=-5020J/mole

w=-5.02kJ/mole

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.