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Heat Capacity

A gas expands...

Question

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

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Solution

Work done $= P Δ V = (3 \times 2) L . a t m = 6 L . a t m = 607.8 J$
where,
$Δ V = 5 - 3 = 2 L$

Heat lost $=$ Heat gained by $H_{2} O = m s Δ t$
$\begin{matrix} \Rightarrow Δ t = \frac{607.8}{10 \times 18 \times 4 \times 4.18} = 0.808 K \Rightarrow T_{f} = 290 + 0.808 = 290.808 K \end{matrix}$

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Similar questions

3.0 {d m}^{3}

5.0 {d m}^{3}

4.184 J g^{- 1} K^{- 1}

3.0 {d m}^{3}

5.0 {d m}^{3}

3.0

10.0

290.0 K

= 4.184 J K^{- 1} g^{- 1}

3 L t o 5 L

3 a t m

290 K

3 d m^{3}

21 d m^{3}

2.1 a t m .

10

290 K .

(i n K)

(Given: Specific heat of water = 4.2 J g^{- 1} K^{- 1}, 1 L . a t m = 101.325 J)

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