Question

A sample consists of 2.00 mole He is expanded isothermally at 22 degree celsius from 22.8 dm³ to 31.7 dm³(a) reversibly, (b) against a constant external pressure equal to the final pressure of the gas, and (c) freely ( against zero external pressure). For the three processes calculate q, w, Delta U, and Delta H.

Answer 1

[AI-generated answer — Bot GPT 3.5]

Given:

Initial moles of He, n = 2.00 mol
Initial volume, V1 = 22.8 dm³
Final volume, V2 = 31.7 dm³
Initial temperature, T = 22°C = 295.15 K

We know that for an ideal gas undergoing isothermal expansion, q = 0 as there is no change in temperature. Also, for an isothermal process:
Delta U = 0 (internal energy remains constant)
Delta H = 0 (enthalpy remains constant)

(a) Reversible isothermal expansion:
Since it is a reversible process, the work done is given by:
w = -nRT * ln(V2/V1)
= -2.00 mol * 8.314 J/molK * 295.15 K * ln(31.7 dm³ / 22.8 dm³)
= -2.00 mol * 8.314 J/molK * 295.15 K * ln(1.3912)
= -2.00 mol * 8.314 J/molK * 295.15 K * 0.3304
= -1641.78 J

(b) Against a constant external pressure:
For an isothermal process, q = -w
q = -(-1641.78 J) = 1641.78 J
Since the external pressure is the final pressure of the gas, work done is equal to the pressure-volume work, so w = -Pext * (V2 - V1)
w = -101.3 kPa * (31.7 dm³ - 22.8 dm³)
w = -101.3 kPa * 8.9 dm³
w = -907.57 J

Delta U = q + w
Delta U = 1641.78 J - 907.57 J = 734.21 J

(c) Freely (against zero external pressure):
Since there is no external pressure to do work against, the work done is zero.
w = 0
q = 0
Delta U = 0
Delta H = 0

In conclusion:
(a) q = 0, w = -1641.78 J, Delta U = 0, Delta H = 0
(b) q = 1641.78 J, w = -907.57 J, Delta U = 734.21 J, Delta H = 0
(c) q = 0, w = 0, Delta U = 0, Delta H = 0