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 $4.184\, Jg^{-1}K^{-1}$ ?

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

Answer 1

Answer: (A)

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