Question

An ideal gas in a baloon is kept in a thermal equilibrium with its constant tempreture surroundings. How much work is done by the gas if the outside pressure is slowly reduced,allowing baloon to expand to 6.0 times its original size? The baloon initially has a pressure of 645.0 Pa and volume of 0.10m

Answer 1

The work of the gas is

$$W=\int^{V_2}_{V_1}PdV.$$

Since we deal with an isothermal process, we can use the ideal gas law to find the relationship between pressure and volume:

$$PV=nRT,\\\space\\ P=\frac{nRT}{V}=\frac CV.$$

Substitute:

$$W=\int^{V_2}_{V_1}PdV=C\int^{V_2}_{V_1}\frac{dV}{V}=C\ln\frac{V_2}{V_1}.$$

Since the amount of substance of gas does not change, we can find the C from the initial pressure and volume:

$$C=P_1V_1:\\\space\\ W=P_1V_1\ln\frac{V_2}{V_1}=645·0.1·\ln6=115.6\text{ J}.$$