Question

A piston chamber filled with ideal gas is kept in a constant-temperature bath at 25.0°C. The piston expands from 25.0 mL to 75.0 mL very, very slowly, as illustrated in the Figure below. If there is 0.00100 mole of ideal gas in the chamber, calculate the work done by the system

Answer 1

Answer : The work done by the system is, 2.2722 J

Explanation :

The expression used for work done in reversible isothermal expansion will be,

$w=nRT\ln (\frac{V_2}{V_1})$

where,

w = work done = ?

n = number of moles of gas  = 0.00100 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = $25^oC=273+25=298K$

$V_1$ = initial volume of gas  = 25 mL

$V_2$ = final volume of gas  = 75 mL

Now put all the given values in the above formula, we get:

$w=0.00100mole\times 8.314J/moleK\times 298K\times \ln (\frac{75}{25})$

$w=2.722J$

Therefore, the work done by the system is, 2.2722 J