Question

A sample consists of 2.0 mole He was expanded isothermally at 22 degree celsius from 22.8 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 3 processes, calculate q, w, Delta U, and Delta H.

Answer 1

[AI-generated answer — Bot GPT 3.5]

Given:

Initial moles of He (n): 2.0 moles
Initial volume (V1): 22.8 dm³
Initial temperature (T1): 22°C
Final temperature (T2): 22°C

First, let's convert the initial volume from dm³ to m³:
V1 = 22.8 dm³ = 0.0228 m³

We know that for an isothermal process:
q = -w

(a) Reversible expansion:
For a reversible isothermal expansion, the work done (w) can be calculated using the formula:
w = -nRT * ln(V2/V1)
w = -2.0 * 8.31 * 295 * ln(V2/0.0228)

Since it is an isothermal process:
q = -w

Delta U = 0 (since it is an isothermal process)

For an isothermal process, Delta H = 0 (since there is no change in temperature)

(b) Against constant external pressure:
Since the external pressure is equal to the final pressure of the gas, the work done is given by:
w = -Pext * (V2 - V1)

Since it is an isothermal process:
q = -w

Delta U = 0 (since it is an isothermal process)

For an isothermal process, Delta H = 0 (since there is no change in temperature)

(c) Freely (against zero external pressure):
Since the external pressure is zero, the work done is given by:
w = -Pext * (V2 - V1)
But since the external pressure is zero, the work done is also zero.

Since no work is done, q = 0

Delta U = 0 (since it is an isothermal process)

For an isothermal process, Delta H = 0 (since there is no change in temperature)