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 $J{g}^{-1}{K}^{-1}$?

• A. 300
• B. 600
• C. 200
• D. 1000

Answer 1

Answer: (A)

300

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