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

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Solution

Work done $= P \times d V = 3.0 \times (5.0 - 3.0)$
$= 6.0 l i t r e - a t m = 6.0 \times 101.3 J$
$= 607.8 J$
Let $Δ T$ be the change in temperature
Heat absorbed $= m \times s \times Δ T$
$= 10.0 \times 18 \times 4.184 \times Δ T$
Given, $P \times d V = m \times s \times Δ T$
or $Δ T = \frac{P \times d V}{m \times s} = \frac{607.8}{10.0 \times 18.0 \times 4.184} = 0.807$
Final temperature $= 290 + 0.807 = 290.807 K$

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