Question

Consider the adiabatic, reversible expansion of a closed 1 mole sample of monatomic ideal gas from P1 = 100 bar, V1 = 1dm3, and T1 = 1200K to V2 = 1.5 dm3.

What is the final temperature of the gas? What are the values of ΔE, ΔS and w for the process described in the previous question? ΔE = kJ ΔS = J/K w = kJ

Answer 1

From 1st Law dU = -pdV ⇒ CvdT = -pdV along path

Cv dT = - pdV and p = RT/V

Cv dT/T = -R dV/V

On integrating both sides from T1 to T2 and V1 to V2 respectively we get as follows

[T2/T1] = [ V1/V2]^Y-1

For monoatomic gas Y = 5/3

Substituting all other values we get as follows

T2/1200 = [1/1.5]^5/3 - 1

T2/1200 = (0.66)^0.66 = 0.76

T2 = 912.17 K

the values of ΔE, ΔS and w for the process described in the previous question are as follows

1 mole gas (V1,T1) = 1 mole gas (V2,T2)

adiabatic ⇒ đq = 0

Reversible ⇒ đw = -pdV = - 100 x 0.5 = - 50 kJ

Ideal gas ⇒ dU = dE = CvdT = - pdV according to first law so dU = 50 kJ

dS = 0 as dq = 0

as ds = ^initial_final dqrev/ T = 0 so dS = 0