Question

Consider a process in which one mole of

a monatomic ideal gas is compressed from

a volume of V1 =1.459m3 to V2 =1m3 at a

constant temperature of T =353.7 K.

(a) What is the entropy change of the gas (in

J/K units)?

(b) What is the change in the value of PV for

the gas (in J units)?

(c) What is the energy change of the gas (in J

units)?

(d) What is the enthalpy change of the gas

(in J units)?

(e) What is the Helmholtz energy change of the

gas (in J units)?

(f) What is the Gibbs energy change of the gas

(in J units)?

23. Perform a Legendre transform of U(S, V,Ni) in

order to obtain a new potential function whose

natural variables are S, P, and Ni (and make

sure to indicate the partial derivative that is

used in order to transform the variable V to P

Answer 1

(a) Formula: S = nR ln(V₂/V₁), where R is universal gas constant = 8.314 J mol^-1 K^-1 and n = no. of moles of gas

Now, S = 1 mol * 8.314 J mol^-1 K^-1 * ln(1 m³/1.459 m³)

= -3.141 J/K

(b) Formula: (PV) = PV

According to the first law of thermodynamics, TS = E + PV

(Note: Since the temperature is constant, the change in internal energy, i.e. E = 0)

i.e. PV = TS

i.e. PV = 353.7 K * -3.141 J/K ~ -1111 J

(c) The energy change of the gas can be considered as the change in internal energy of the gas (E).

Formula: E = nT

Note: Here, temperature is constant, i.e. T = 0

i.e. E = 1 mol * 0 K = 0 J

(d) ​The enthalpy change of the gas, H = TS + VP

Note: According to Gay-Lussac law, P∝​T, so pressure is also constant, i.e. P = 0

i.e. H = -1111 J + 0 = -1111 J

(e) The Helmholtz energy change of the gas, A = -ST - PV

i.e. A = 0 - (-1111 J) = 1111 J

(f) The Gibbs energy change of the gas, G = -ST + VP

i.e. G = 0 + 0 = 0 J