Answer to Question #91623 in Organic Chemistry for Lid

Question

A 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 ∆E of the process?

Answer 1

Solution.

Since during an isothermal process the change in the internal energy of the system is 0, then in the next process (isobaric) it is not equal to 0. From the first process (isothermal) we find p2.

Boyle-Mariotte law:

$\frac{p1}{p2} = \frac{V2}{V1}$

$p2 = \frac{p1 \times V1}{V2}$

p2 = 10132.5 Pa

$A = p2 \times \Delta V$

A = 19.1925 J

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

$\nu = \frac{V}{Vm}$

$\nu = \frac{1}{22.4} = 0.045 mole$

Q = 188.8448 J

$Q = A + \Delta E$

$\Delta E = Q - A$

$\Delta E = 169.6523 J$

Answer:

$\Delta E = 169.6523 J$