Question

a) One mol of helium gas (considered ideal) undergoes reversible isothermal expansion from 20 dm3 to 70 dm3 at 298 K. Compute ∆U, w and q.
(b) If the gas in part (a) undergoes irreversible into vacuum, compute ∆U, w and q.
(c) Show that for helium gas use in part (a) and (b),

(∂U/∂V)_P=(C_V P)/R

Answer 1

Answer:

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Answer 2

Given:

$v_1 = 20 dm^3\\v_2 = 70 dm^3\\T = 298 K$

No of moles of He, n = 1 mol

To Find:

(a) Internal Energy, $\triangle U$, work interaction, $\delta w$ and heat interaction, $\delta q$

(b) Internal Energy, $\triangle U$, work interaction, $\delta w$ and heat interaction, $\delta q$ during irreversible expansion into vacuum

(c) Show that:- $(\frac{\delta U}{\delta V})_P = \frac{c_v\times P}{R}$

Solution:

(a)

$v_1 = 20 dm^3\\ = 0.02m^3\\v_2 = 70 dm^3\\= 0.07 m^3\\T = 298 K$

We know that for an ideal gas, internal energy, U is a function of temperature only. Since it is an isothermal expansion, therefore, $\triangle T = 0$ and hence $\triangle U = 0$.

Also, we know that work done in an isothermal process,$\delta w= p_1\times v_1 \ln \frac{v_2}{v_1}$

and for ideal gas, $p_1\times v_1 = n\times R\times T$

Substituting the values in above equations we get,

$p_1\times v_1 = n\times R\times T\\\\p_1 =\frac{ 1\times 8.314\times 298}{0.02}\\\\p_1= 123878.6 \frac{N}{m^2}$

Now,

$\delta w= 123878.6\times 0.02 \ln \frac{0.07}{0.02}\\\\\delta w = 3103.81 J$

We also know as per 1st Law of Thermodynamics that

, $\delta q= \triangle U + \delta w \\therefore, \delta q = \delta w = 3103.81 J , since \triangle U = 0$

(b) For, irreversible free expansion, work done is zero.

Also, change in internal energy is zero.

(c) We know that, $\triangle U =c_v dt$

Also, $p_1\times v_1 = n\times R\times T$

On substituting and re arranging the terms, for constant pressure process, we get $(\frac{\delta U}{\delta V})_P = \frac{c_v\times P}{R}$