Question

A gas expands from a volume of $3.0d{m}^3$ to $5.0d{m}^3$ against a constant pressure of $3.0atm$ the work done during expansion is used to heat $10.0$ mole of water of temperature $290.0K$ Calculate the final temperature of water (specific heat of water $=4.184J{K}^{-1}{g}^{-1}$)

• A. $290K$
• B. $289.2K$
• C. $290.8K$
• D. $291.8K$

Answer 1

The correct option is C $290.8K$

$\text{Work done}=P\times dV\text{Work done}=3.0\times (5.0-3.0)=6.0litreatm=6.0\times 101.3J=607.8J$
Let $\Delta T$ be the change in temperature,
$\text{Heat absorbed}=m\times s\times \Delta T=10.0\times 18\times 4.184\times \Delta TP\times dV=m\times s\times \Delta T$

or $\Delta T$ $=\frac{P\times dV}{m\times s}=\frac{607.8}{10.0\times 18.0\times 4.184}=0.807$
Thus, Final temperature $=290+0.807=290.807K$