Question

Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until total volume is 10 litres. How much heat is absorbed and how much work is done in expansion?

Answer 1

Heat absorbed for an isothermal expansion is given as, `nRT ln((V_2)/(V_1))` where, `n` is the number of moles,`T` is the constant temperature associated with the process, `V_2` is the final volume and `V_1` was the initial volume.

Considering, `n=1` applying ideal gas law,we can find the temperature,

so,`P_1V_1=nRT`

or, `T=(P_1V_1)/(nR)`

or, `RT=P_1V_1` for,`n=1`

So,`RT=10*2=20`

So,heat absorbed is `1*20*ln(10/2)=32.2 J`

And we know,from `1` st law of thermodynamics,change in internal energy for an isothermal process is zero,so work done=heat absorbed=`32.2J`