1
Question
Two mole of ideal diatomic gas (Cv,m=52 R) at 300 K and 5 atm expanded irreversly and adiabatically to a final pressure of 2 atm against a constant pressure of 1 atm. Calculate q, w, ΔH and ΔU.
Two mole of ideal diatomic gas (Cv,m=52 R) at 300 K and 5 atm expanded irreversly and adiabatically to a final pressure of 2 atm against a constant pressure of 1 atm. Calculate q, w, ΔH and ΔU.
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Solution
The correct option is C w=−1246 J,q=0,ΔH=−1746 J,ΔU=−1247 J
For Adiabatic irreversible process
nCv(T2−T1)=−pextnR[T2p2−T1P1] ...(1)
Where, Cv=52R
T1=300 K
Pext=1 atm P2=2atm
P1=5 atm
2×52R(T2−300)=−1×2R[T22−3005]
=5(T2−300)=−(T2−120)
=5T2−1500=−T2+120
⇒T2=270 k
Also we know,
ΔU=q+w=w ∵q=0
The workdone in an adiabatic process= w=nCv△T
=2×52×8.314(270−300)
=−1247.1 J
ΔU=w=−1247.1 J
ΔH=△U+nRΔT
=−1247.1+2×8.314(270−300)
⇒ΔH=−1745.9 J
For Adiabatic irreversible process
nCv(T2−T1)=−pextnR[T2p2−T1P1] ...(1)
Where, Cv=52R
T1=300 K
Pext=1 atm P2=2atm
P1=5 atm
2×52R(T2−300)=−1×2R[T22−3005]
=5(T2−300)=−(T2−120)
=5T2−1500=−T2+120
⇒T2=270 k
Also we know,
ΔU=q+w=w ∵q=0
The workdone in an adiabatic process= w=nCv△T
=2×52×8.314(270−300)
=−1247.1 J
ΔU=w=−1247.1 J
ΔH=△U+nRΔT
=−1247.1+2×8.314(270−300)
⇒ΔH=−1745.9 J
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