Question

1 mole of gas occupying 3 litre volume is expanded against a constant external pressure of 1 ATM to a volume of 15 litre the work done by a gas is equal to

Answer 1

Work done = -P∆V

= - 1(15-3)

= -1×12

= -12 L-atm

= -1215.6 Joule.

Answer 2

Answer : The work done by a gas is equal to -1215.6 J

Solution :

This is the case of isothermal irreversible expansion of gas.

Formula used :

$w=-P_{ext}\times (V_2-V_1)$

where,

w = work done by the gas

$P_{ext}$ = external pressure of gas = 1 atm

$V_1$ = initial volume of gas = 3 L

$V_2$ = final volume of gas = 15 L

Now put all the given values in the above formula, we get the work done.

$w=-(1atm)\times (15L-3L)$

$w=-12Latm$

$w=-1215.6J$

conversion : (1 L atm = 101.325 J)

Therefore, the work done by a gas is equal to -1215.6 J