Question

A gas expands from $3 \, dm^{3}$ to $5 \, dm^{3}$ against a constant pressure of 3 atm. The work done during expansion is used to heat 10 mole of water at temperature 290 K. Calculate final temperature of water. Specific heat of water = 4.184 J/ g/K.
[Given: 1 atm = 101.33 J]

• A. $290.81K$
• B. $290.61K$
• C. $290.41K$
• D. $290.21K$

Answer 1

Answer: (A)

Work is done against constant P
$\Delta V=5-3=2 \, dm^{3}=2 \, litre \, ;P=3 \, atm$
$\therefore W=-P\Delta V$
$=-3\times 2 \, litre \, atm$
$= - 6 \times 101.33 = - 608 \text{ J}$
Now, this work is used up in heating water
$\therefore \left|\text{W}\right| = \text{n} \times \text{C} \times \text{ΔT}$
$608=10\times 4.184\times 18\Delta T$
$\therefore \Delta T=0.81$
$\therefore Final \, temperature=T_{1}+\Delta T$
$=290+0.81=290.81 \, K$