Two moles of a diatonic gas (gama=1.4) At 127°C are expanded adiabatically to twice the original volume. Calculate
i) The final temperature and
ii) The work done in this process
given, R=8.3J/K/mole and (0.5)^0.4 = 0.76
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process is adiabatically , so
PV^Y = constant
also, V^(Y-1)T =constant
so, {V1/V2}^(Y -1) =(T2/T1)
given, T1 =127°C = 400K
V2/V1 = 2
now,
(1/2)^(1.4-1) =(T2/400)
400(1/2)^(0.4) = T2
T2 = 304 K
2) work done = (P1V1 -P2V2)/(Y-1)
=nR(T1-T2)/(Y-1)
=2 × 8.3 × (400-304)/(0.4)
=2 × 8.3 × 96/0.4 Joule
= 3984 Joule
PV^Y = constant
also, V^(Y-1)T =constant
so, {V1/V2}^(Y -1) =(T2/T1)
given, T1 =127°C = 400K
V2/V1 = 2
now,
(1/2)^(1.4-1) =(T2/400)
400(1/2)^(0.4) = T2
T2 = 304 K
2) work done = (P1V1 -P2V2)/(Y-1)
=nR(T1-T2)/(Y-1)
=2 × 8.3 × (400-304)/(0.4)
=2 × 8.3 × 96/0.4 Joule
= 3984 Joule
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