Question

A gas expands from a volume of $3.0{dm}^3$ to $5.0{dm}^3$ against a constant external pressure of 3.0 atm. The work done during of temperature 290.0 K. Calculate the final temperature of water (specific heat of water=$4.184J{g}^{-1}{K}^{-1}$)

Answer 1

$W_{PV}=-P\Delta V=-3(5-3)=-6$

$L-atm=-6\times 101.325=-607.95J$

Molar heat capacity of water

$=4.184\times 18=75.312J/mol$

Now$,$ $Q=ms\Delta T$

or $\Delta T=\frac{Q}{ms}=\frac{607.95}{\left(10\times 75.312\right)}=0.807$

Final temperature $=290+0.807=290.807K$

We get$,$ $290.807K$