Question

61N 16. Two moles of an ideal gas are compressed in a cylinder at a constant temperature of 65 °C until the original pressure has tripled. (a) Sketch a pV diagram for this process. (b) Calculate the amount of work done.

Answer 1

a) The process is isothermal

Therefore,

P1 V1 = P2 V2

Given, P2 = 3 P1

P1 V1 = 3 P1 V2

implies, V1 = 3V2 (gass is compressed)

7 338 K P co, נא 2 V ףw

(b) Work done when pressure is changed is given by,

$W = \int_{v1}^{v2} P dV$

From the ideal gas law, $P = \frac{nRT}{V}$

Therefore,

v2 nRT AP-

$= nRT \int_{v1}^{v2}\frac{1}{V} dV$

$= nRT [ln \; V]_{v1}^{v2} \\ W = nRT ln \frac{V_2}{V_1} \\ V_1 = 3 V_2 \\ W = nRT ln (\frac{V_2}{3V_2}) \\ = nRT ln (1/3) \\ n = 2\\ R = 8.314 J/mol K \\ T = 65 \degree C = 338 K\\ W = 2 \times 8.314 \times \ 338 \times ln (1/3) J \\ = -6174.49 \; J \\$