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Standard XII

Physics

1st Law of Thermodynamics

1 g of H2O ...

Question

$1 g$ of $H_{2} O$ changes from liquid to vapour phase at constant pressure at $1 a t m$ . The volume increases from $1 c c t o 1671 c c$ . The heat of vaporisation at this pressur is $540 c a l / g$ . The increase in internal energy of water is

2099 J

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3000 J

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992 J

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2122 J

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Solution

The correct option is A $2099 J$
Work done is given as-
$W = P (d V) = 1.01 \times 10^{5} (1671 - 1) \times 10^{- 6} = 167 J$
$Δ Q$ = $Δ U$ + $Δ W$
or
$Δ U$ = $Δ Q$ - $Δ W$
$= m L - 167 = (540 \times 4.2) - 167 = 2099 J$

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1 g of H2O changes from liquid to vapour phase at constant pressure at 1 atm. The volume increases from 1 cc to 1671 cc. The heat of vaporisation at this pressur is 540 cal/g. The increase in internal energy of water is

The correct option is A 2099JWork done is given as-W=P(dV)=1.01×105(1671−1)×10−6=167JΔQ = ΔU + ΔWor ΔU = ΔQ - ΔW=mL−167=(540×4.2)−167=2099J

$1 g$ of $H_{2} O$ changes from liquid to vapour phase at constant pressure at $1 a t m$ . The volume increases from $1 c c t o 1671 c c$ . The heat of vaporisation at this pressur is $540 c a l / g$ . The increase in internal energy of water is

The correct option is A $2099 J$
Work done is given as-
$W = P (d V) = 1.01 \times 10^{5} (1671 - 1) \times 10^{- 6} = 167 J$
$Δ Q$ = $Δ U$ + $Δ W$
or
$Δ U$ = $Δ Q$ - $Δ W$
$= m L - 167 = (540 \times 4.2) - 167 = 2099 J$