Question

A gas expands from $3d{m}^3$ to $21d{m}^3$ against a constant pressure of $2.1atm.$ The work during expansion is used to heat $10$ mole of water of temperature $290K.$ Final temperature of water $\left(inK\right)$is:
(Round the answer to the nearest intiger.)
$(\text{Given: Specific heat of water}=4.2J{g}^{-1}{K}^{-1},1L.atm=101.325J)$

Answer 1

Work is done against constant $P$.
$\Delta V=21-3=18d{m}^3=18litreP=2.1atm\therefore W=-P.\Delta V=-2.1\times 18litreatm=-2.1\times 18\times 101.325J$
Now this work is used up in heating water.
$W=m\times C\times \Delta T\Rightarrow 2.1\times 18\times 101.325=10\times 18\times 4.2\Delta T\Rightarrow \Delta T=5.06K\therefore \text{Final temperature}=T_1+\Delta T=290+5.06=295.06K\approx 295K$