An ideal gas of 0.1 kmol at the initial state of 298.15K and 303.9kPa occupies o

Question

An ideal gas of 0.1 kmol at the initial state of 298.15K and 303.9kPa occupies one chamber of an insulated composite system. The other chamber (of double volume of the first) contains a vacuum. Determine the volume of the first chamber V1. After the removal of the partition between the two chambers, the ideal gas undergoes an adiabatic free expansion (see Problem 4.2) from its initial volume V1 to its final volume 3 V1. Explain why the gas remains at the same temperature at its final state (hint: note the values of heat exchange Q and work W in this case; give precise reason for the W value you assume). And determine the entropy change of the free expansion.

Explanation / Answer

a) using Ideal gas law

PV = nRT

( 303.9 ) x 10^3 ( V) = 0.1 x 10^3 ( 25/3) (298.15)

V 1 = 1.223 m^3 -=1233 liters

b) Adibatic free expansion,volume of the chamber doesn't change ( np work is done as w= pdv) , and further as the system is insulated, dQ = 0

dU = W + dQ

dU = 0 + 0 = 0 , the internal energy of the gas doesn't change during the expansion,which implies that temp doesn't change.

U = 3/2 nRT ( for monoatomic gas)

U = kT ( where K = constant = 3/2nR)

dIU = kdT = 0

which implie sthat final temp and initial temp is same.

c) S ( entropy change) =nR ln ( Vf/Vi)

S = ( 0.1 x 10^3) ( 25/3) ln ( 3/1) = 915.51 J/K

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