A sample consisting of 1.00 mol of perfect gas molecules at 27 °C is expanded isothermally from an initial pressure of 3.00 atm to a final pressure of 1.00 atm in two ways: (a) reversibly, and (b) against a constant external pressure of 1.00 atm. Evaluate q, w, ΔU, ΔH, ΔS, ΔSsurr, and ΔStot in each case.

a) reversibly

ΔU = 0

q = 2740.16 J

w = - 2740.16 J

ΔH = 0

ΔS (total) = 0

ΔS (sys) = 9.13 J/K

ΔS (surr) = - 9.13 J/K

b) against a constant external pressure of 1.00 atm

ΔU = 0

w = - 1.66 kJ

q = 1.66 kJ

ΔH = 0

ΔS (sys) = 9.13 J/K

ΔS (surr) = - 5.543 J/K

ΔS (total) = 3.587 J/K

Explanation:

Step 1: Data given:

Number of moles = 1.00 mol

Temperature = 27.00 °C = 300 Kelvin

Initial pressure = 3.00 atm

Final pressure = 1.00 atm

The gas constant = 8.31 J/mol*K

(a) reversibly

Step 2: Calculate work done

For ideal gases ΔU depends only on temperature. So as it is an isothermal (T constant).

Since the temperature remains constant:

ΔU = 0

ΔU = q + w

q = - w

w = - nRT ln (Pi/Pf)

⇒ with n = the number of moles of perfect gas = 1.00 mol

⇒ with R = the gas constant = 8.314 J/mol*K

⇒ with T = the temperature = 300 Kelvin

⇒ with Pi = the initial pressure = 3.00 atm

⇒ with Pf = the final pressure = 1.00 atm

w = - 1*8.314 * 300 * ln (3)

w = - 2740.16 J

q = - w

q = 2740.16 J

Step 3: Calculate change in enthalpy

Since there is no change in energy, ΔH = 0

Step 4: Calculate ΔS

for an isothermal process

ΔS (total) = ΔS (sys) + ΔS (surr)

ΔS (sys) = - ΔS (surr)

ΔS (sys) = n*R*ln (pi/pf)

ΔS (sys) = 1.00 * 8.314 * ln (3)

ΔS (sys) = 9.13 J/K

ΔS (surr) = - 9.13 J/K

ΔS (total) = ΔS (sys) + ΔS (surr) = 0

(b) against a constant external pressure of 1.00 atm

Step 1: Calculate the work done

w = - Pext*ΔV

w = - Pext * (Vf - Vi)

⇒ with Vf = the final volume

⇒ with Vi = the initial volume

We have to calculate the final and initial volume. We do this via the ideal gas law P*V=n*R*T

V = (n*R*T) / P

Initial volume = (n*R*T) / Pi

⇒ Vi = (1*0.08206 * 300) / 3

⇒ Vi = 8.206 L

Final volume = (n*R*T) / Pf

⇒ Vf = (1*0.08206 * 300) / 1

⇒ Vf = 24.618 L

The work done w = - Pext * (Vf - Vi)

w = - 1.00 * (24.618 - 8.206)

w = - 16.412 atm*L

w = - 16.412 * (101325/1atm*L) * (1kJ/1000J)

w = - 1662.9 J = - 1.66 kJ

Step 2: Calculate the change in internal energy

ΔU = 0

q = - w

q = 1.66 kJ

ΔH = 0 because there is no change in energy

Step 3: Calculate ΔS

ΔS (sys) = n*R*ln (3)

ΔS (sys) = 1.00 * 8.314 * ln (3)

ΔS (sys) = 9.13 J/K

ΔS (surr) = - q/T

ΔS (surr) = - 1662.9J/300K

ΔS (surr) = - 5.543 J/K

ΔS (total) = ΔS (surr) + ΔS (sys) = - 5.543 J/K + 9.13 J/K = 3.587 J/K