REVERSIBLE EXPANSION OF 0.5 mol OF AN IDEAL GAS Wmax IS -3.918KJ.THE VALUE OF ΔU IS
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Solution:-
Work done in an isothermal reversible process is given by-
W=−nRTln
V
1
V
2
Where,
n = no. of moles =1
R= molar gas constant =8.314Jmol
−1
K
−1
T= temperature =300K
V
1
=10dm
3
V
2
=20dm
3
∴W=−1×8.314×300×ln
10
20
⇒W=−2494.2×(ln2−ln1)=−1728.8 J=−1.73 kJ
Now, from first law of thermodynamics,
ΔU=ΔH+W
Since temperature is constant, i.e. ΔU=0.
∴ΔH=−W=−(−1.73)=+1.73KJ
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