Question

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

• A. $300$
• B. $600$
• C. $200$
• D. $1000$

Answer 1

Answer: (A)

Work done, $w=-P_{ext.}dV$
$W=-2\times 2.5=-5Latm=-506.3J$
Because this work is used in raising the temperature of water, so work done is equal to the heat supplied i.e.,
$w=q=m\cdot c_s\cdot s\cdot \Delta T$
Given that, $m=18g(=1$ mole$)$, $c_s=4.184J{g}^{-1}{K}^{-1}$,
$q=+506.3J$, $\Delta T=?$ (Heat is given to water)
$\Delta T=\frac{q}{c_s\cdot m}$
$=\frac{506.3}{4.184\times 18}=6.72$
$\therefore$ Final temperature, $T_f=T_i+\Delta T$
$=293+6.72$
$=299.72K\approx 300K$