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Standard XII

Chemistry

Enthalpy

1 gm of water...

Question

1 gm of water at a pressure of $1.01 \times 10^{5}$ Pa is converted into steam without any change of temperature. The volume of 1 g of steam is 1671 cc and the latent heat of evaporation is 540 cal. The change in internal energy due to evaporation of 1 gm of water is

\approx

167 c a l

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\approx

500 c a l

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540 c a l

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581 c a l

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Solution

The correct option is B $\approx$ $500 c a l$
$d W = P$ $Δ V$ = $1.01 \times 10^{5} [1671 - 1] \times 10^{- 6}$ joule
= $\frac{1.01 \times 167}{4.2}$ cal
= 40 cal. nearly
$Δ Q = m L = 1 \times 540$ ,
$Δ Q = Δ W + Δ U$
or $Δ U$ $= 540 - 40$
$= 500 c a l .$

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Similar questions

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1 gm of water at a pressure of 1.01×105 Pa is converted into steam without any change of temperature. The volume of 1 g of steam is 1671 cc and the latent heat of evaporation is 540 cal. The change in internal energy due to evaporation of 1 gm of water is

The correct option is B ≈ 500caldW=PΔV = 1.01×105[1671−1]×10−6 joule= 1.01×1674.2cal= 40 cal. nearlyΔQ=mL=1×540,ΔQ=ΔW+ΔUor ΔU =540−40 =500cal.

1 gm of water at a pressure of $1.01 \times 10^{5}$ Pa is converted into steam without any change of temperature. The volume of 1 g of steam is 1671 cc and the latent heat of evaporation is 540 cal. The change in internal energy due to evaporation of 1 gm of water is

The correct option is B $\approx$ $500 c a l$
$d W = P$ $Δ V$ = $1.01 \times 10^{5} [1671 - 1] \times 10^{- 6}$ joule
= $\frac{1.01 \times 167}{4.2}$ cal
= 40 cal. nearly
$Δ Q = m L = 1 \times 540$ ,
$Δ Q = Δ W + Δ U$
or $Δ U$ $= 540 - 40$
$= 500 c a l .$