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Heat of Reaction

A gas expands...

Question

A gas expands from $3 d m^{3}$ to $21 d m^{3}$ against a constant pressure of $2.1 a t m .$ The work during expansion is used to heat $10$ mole of water of temperature $290 K .$ Final temperature of water $(i n K)$ is:
(Round the answer to the nearest intiger.)
$(Given: Specific heat of water = 4.2 J g^{- 1} K^{- 1}, 1 L . a t m = 101.325 J)$

295

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295.00

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295.0

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Solution

Work is done against constant $P$ .
$Δ V = 21 - 3 = 18 d m^{3} = 18 l i t r e P = 2.1 a t m ∴ W = - P . Δ V = - 2.1 \times 18 l i t r e a t m = - 2.1 \times 18 \times 101.325 J$
Now this work is used up in heating water.
$W = m \times C \times Δ T \Rightarrow 2.1 \times 18 \times 101.325 = 10 \times 18 \times 4.2 Δ T \Rightarrow Δ T = 5.06 K ∴ Final temperature = T_{1} + Δ T = 290 + 5.06 = 295.06 K \approx 295 K$

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Similar questions

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)

d m^{3}

d m^{3}

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

3.0 d m^{3}

5.0 d m^{3}

3.0 a t m

10.0

290.0 K

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

3.0 {d m}^{3}

5.0 {d m}^{3}

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

3 L t o 5 L

3 a t m

290 K

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