Question
Temperature of $$5\,moles$$ of a gas is decreased by $$2K$$ at constant pressure. Indicate the correct statement
A.
Work done by gas is $$ = 5R$$
B.
Work done over the gas is $$ = 10\,R$$
C.
Work done by the gas $$ = 10\,R$$
D.
Work done $$ = 0$$
Answer :
Work done over the gas is $$ = 10\,R$$
Solution :
For $$5\,moles$$ of gas at temperature $$T,$$
$$P{V_1} = 5RT$$
For $$5\,moles$$ of gas at temperature $$T - 2,$$
$$P{V_2} = 5R\left( {T - 2} \right)$$
$$\eqalign{
& \therefore \,\,P\left( {{V_2} - {V_1}} \right) \cr
& = 5R\left( {T - 2 - T} \right); \cr
& P\Delta V = - 10R, \cr
& - P\Delta V = 10R \cr} $$
When $$\Delta V$$ is negative, $$W$$ is $$ + ve.$$