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Question

Two moles of an ideal monoatomic gas at NTP are compressed adiabatically and reversibly to occupy a volume of 4.48 dm3. Calculate the amount of work done, ΔU, final temperature and pressure of the gas. CV for ideal gas =12.45 JK1 mol1

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Solution

For an ideal gas, γ=CPCV=1.667
Initial volume, V1=2×22.4=448 dm3
Initial pressure, P1=1 atm
Initial temperature, T1=273 K
Final volume, V2=4.48 dm3
Let the final pressure be P2 and temperature be T2.
Applying P1V1γ=P2V2γ
P1P2=(V2V1)γ=(4.4844.8)1.667
or P2P1=(10)1.667
P2=(10)1.667(P1=1 given)
logP2=1.667log10=1.667
P2=antilog1.667=46.45 atm
Final temperature=P2V2P1V1.T1=46.45×4.481×44.8×273=1268K
Work done on the system=n.CV.ΔT
=2×12.45×(1268273)
=2×12.45×995=24775.5 J
From the first law of thermodynamics,
ΔU=q+w=0+24775.5=24775.5 J

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