An ideal gas does work on its surroundings when it expands by 2.5 L against external pressure 2 atm

Question

An ideal gas does work on its surroundings when it expands by 2.5 L against external pressure 2 atm . This work done is used to heat up 1 mole of water at 293 K . What would be the final temperature of water in Kelvin if specific heat for water is $4.184 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}$ ?\

Professor PC Thomas Classes · Answer

a) : Work done, $w=-P_{	ext {ext. }} d V$
$
w=-2 	imes 2.5=-5 \mathrm{~L} \mathrm{~atm}=-506.3 \mathrm{~J}
$
Because this work is used in raising the temperature of water, so work done is equal to the heat supplied i.e.,
$
w=q=m \cdot c_s \cdot \Delta T
$
Given that, $m=18 \mathrm{~g}$ (= 1 mole), $c_s=4.184 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}$, $q=+506.3 \mathrm{~J}$ (Heat is given to water), $\Delta T=$ ?
$
\Delta T=\frac{q}{c_s \cdot m}=\frac{506.3}{4.184 	imes 18}=6.72
$
$	herefore \quad$ Final temperature, $T_f=T_i+\Delta T=293+6.72$
$
=299.72 \mathrm{~K} \approx 300 \mathrm{~K}
$