Question

1. 1L sample of gas at 1atm pressure and 298K expands isothermally to 10L. It is then heated to 500K, compressed to 1L and then cooled 298K. What is the ∆U of overall the process?

Answer 1

Solution.

The first process proceeds isothermically (at T=const), therefore, the change in internal energy is equal to 0. From the first process, we need to find out what pressure was formed by the SB using the klayperon-Mendeleev equation.

$p1 \times V1 = p2 \times V2$

p2 = 0.1 atm.

$p \times V = n \times R \times T$

$n = \frac{p \times V}{R \times T}$

n = 0.041 mol

The second process clearly occurs at constant pressure (p=const). Expression of the first law of thermodynamics for an Isobaric process:

$\Delta U = Q-A$

$A = -p \times \Delta V$

$A = -0.1*101325*(0.01-0.001) = -91.19 \ J$

$Q = \nu \times (\frac{5}{2}) \times R \times \Delta T$

$Q = 0.041 \times (\frac{5}{2}) \times 8.31 \times 202 = 172.06 \ J$

$\Delta U = 172.06 +91.19 = 263.25 \ J$

Answer:

$\Delta U = 263.25 \ J$