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# A gas expands from a volume of $3\: dm^3$ against a constant external pressure of $3\: atm$. to a volume of $5\: dm^3$. The work of expansion is used to heat $10\: ml$ of water at $290\: K$. Calculate the final temperature of the water.

$\begin {array} {1 1} (A)\;289.193 \: K & \quad (B)\;-289.193 \: K \\ (C)\;290.807\: K & \quad (D)\;-290.807 \: K \end {array}$

$W_{PV} = P \Delta V = -3(5-3) = -6\: L-atm = -6 \times101.325 = -607.95 \: J$
Molar heat capacity of water $= 4.184 \times 18 = 75.312\: J/mol$
Now, $Q = ms \Delta T\: \: \: \: \: or, \: \Delta T = \large\frac{Q}{ms}$$= \large\frac{607.95}{(10 \times 75.312)}$$ = 0.807$
Final temperature =$290 + 0.807 = 290.807 \: K$