Question

One mole of an ideal gas is compressed at a constant temperature of 55 ^oC from 16.5 L to 12.8 L using a constant external pressure of 1.6 atm. Calculate w, q, ΔH and ΔS for this process.

w = (?) kJ

q = (?) kJ

ΔH = (?) kJ

ΔS = (?) J/(mol*K)

Answer 1

work done in isothermal process W - - Pext dV

volume V1= 16.5 L;V2 = 12.8 L
Pext = 1.6 atm

So W = - 1.6 atm[12.8L-16.5 L] =
W= 5.92 J
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Positive workdone indicate work is on the system by the surrounding

Internal energy depends on temperature,and here the change in temperature

dT =0
So dU =0

WE know dU = q +w
when dU=0
q=- w for isothermal process

we have w= 5.92 j

then q= -5.92
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For isothermal process ,enthalpy change is zero

deltaH = 0
***********

delta S for isothermal process
dS = nRln[V2/V1]

n= 1 mole of gas
volume V1= 16.5 L;V2 = 12.8 L

R= 8.314 J/ K-mol

dS = 1*8.314 J/mol-K ln[12.8 l/16.5 l] = -2.11 J/K-mol

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