For an adiabatic process, the change in T is determined by the change in V. In this problem you will compute the contributions to S from the V and T terms separately, then add them up to find the total entropy change for an adiabatic process. Argon gas, initially at pressure 100 kPa and temperature 300 K, is allowed to expand adiabatically from 0.01 m3 to 0.026 m3 while doing work on a piston.

For an adiabatic process, the change in T is determined by the change in V. In this problem you will compute the contributions to S from the V and T terms separately, then add them up to find the total entropy change for an adiabatic process.

Argon gas, initially at pressure 100 kPa and temperature 300 K, is allowed to expand adiabatically from 0.01 m³ to 0.026 m³ while doing work on a piston.

1) What is the change in entropy due to the volume change alone, ignoring any effects of changing internal energy? ΔS = ? J/K

2) For this adiabatic expansion, what is the final temperature? T[tex]_f[/tex] = ? K

Answer:

1) the change in entropy due to the volume change alone, ignoring any effects of changing internal energy is 3.185 J/K.

2) the final temperature is 158.66 K

Explanation:

Given the data in the question;

ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])

P[tex]_i[/tex] = 100 kPa = 100000 Pa

V[tex]_i[/tex] = 0.01 m³

V[tex]_f[/tex] = 0.026 m³

T[tex]_i[/tex] = 300 K

1) the change in entropy due to the volume change alone

from the question; ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])

so change in entropy due to the volume change alone is;

ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex])

we know that, from ideal gas law; PV = nRT

so, nR = P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] ---- let this be equation 1

∴ ΔS = P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] × ln(V[tex]_f[/tex]/V[tex]_i[/tex])

we substitute

ΔS = [( 100000 Pa × 0.01 m³) / 300 K ] × ln(0.026m³ / 0.01m³ )

ΔS = 3.185 J/K

Therefore, the change in entropy due to the volume change alone, ignoring any effects of changing internal energy is 3.185 J/K.

2) Final temperature

we know that, in an adiabatic expansion;

[tex]PV^Y[/tex] = K

where Y = 5/3

so

[tex]P_i[/tex][tex]V_i^{(5/3)[/tex] = [tex]P_f[/tex][tex]V_f^{(5/3)[/tex]

[tex]P_f[/tex] = [tex]P_i[/tex][tex]( \frac{V_i}{V_f})^{(5/3)[/tex]

we substitute

[tex]P_f[/tex] = ( 100000 Pa) [tex]( \frac{0.01 m^3}{0.026 m^3})^{(5/3)[/tex]

[tex]P_f[/tex] = 20341.255 Pa

Also from ideal gas law;

PV = nRT

T = PV / nR

so

T[tex]_f[/tex] = P[tex]_f[/tex]V[tex]_f[/tex] / nR

but from equation 1; nR = PV/T

so

T[tex]_f[/tex] = (P[tex]_f[/tex]V[tex]_f[/tex]) / (P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] )

T[tex]_f[/tex] = ( P[tex]_f[/tex]V[tex]_f[/tex]T[tex]_i[/tex] / P[tex]_i[/tex]V[tex]_i[/tex] )

we substitute

T[tex]_f[/tex] = ( 20341.255 Pa × 0.026 m³ × 300 K) / 100000 Pa × 0.01 m³ )

T[tex]_f[/tex] = 158.66 K

Therefore, the final temperature is 158.66 K