What is value of U heat change at constant volume for reversible isothermal evaporation of 90 g water at 100- C- Assuming water vapour behaves asan ideal gas and - H vap water -540 cal g -1A- 9 - 103 kcalB- 6 - 103 kcalC- 4-49 - 103 calD- none of these

The correct option is C 4.49 × nbsp; 10^3 calHeat of system ( H)_v = m × L_v = 90 × 540 = 48600 cal H = U + P(V_V V_L) = U + nRT or, nbsp; nbsp; ...

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Standard XII

Chemistry

Heat of Reaction

What is value...

Question

What is value of $△ U$ (heat change at constant volume) for reversible isothermal evaporation of 90 g water at $100^{o} C .$ Assuming water vapour behaves as an ideal gas and $△ H_{v a p})_{w a t e r} = 540 c a l g^{- 1}$

9 \times 10^{3} k c a l

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6 \times 10^{3} k c a l

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4.49 \times 10^{3} c a l

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none of these

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Solution

The correct option is C $4.49 \times 10^{3} c a l$
Heat of system $(△ H)_{v} = m \times L_{v}$
$= 90 \times 540 = 48600 c a l$
$△ H = △ U + P (V_{V} - V_{L}) = △ U + n R T$
or, $48600 = △ U + \frac{90}{18} \times 2 \times 373$
or, $△ U = 48600 - 3730 = 44870 c a l$

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Similar questions

Q. What is value of

△ U

(heat change at constant volume) for reversible isothermal evaporation of 90 g water at

100^{o} C .

Assuming water vapour behaves as an ideal gas and

(△ H_{v a p})_{w a t e r} = 540 c a l g^{- 1}

Q. For the reversible isothermal evaporation of 90.0 g of water at

100^{\circ} C

. The value of

Δ U

for the reaction is:

[Assuming that water vapours behave as an ideal gas and heat of evaporation of water is

540 c a l g^{- 1}

and

R = 2 c a l m o l^{- 1} K^{- 1}

]

Q. The values of

Δ H

and

Δ U

for the reversible isothermal evaporation of 90.0 g of water at

100^{o}

C are

x

c a l

y

c a l

. Assume that water vapour behaves as an ideal gas and heat of evaporation of water is 540 cal

g^{- 1}

(R = 2 c a l {m o l}^{- 1} K^{- 1})

. The value of

x + y

is:

Determine the value of ∆H and ∆U for the reversible isothermal evaporation of 90.0g of water at 100°C. Assume that water vapour behave as am ideal gas and heat of eveporation of water is 540cal.

MY PROBLEM

Only tell me How ∆H is calculated here, because if I know∆H then I can use the equation

∆H = ∆U + nRT

In my book it is given that

∆H = 90.0 x 540

= 48600 cal

But I want to know that why they have multipled the mass to heat of eveporation to find out ∆H

Heat of reaction for $C_{6} H_{12} O_{6} (s) + 6 O_{2} \to 6 C O_{2} (g) + 6 H_{2} O (v)$ at constant pressure is $- 651 K c a l$ at $17^{\circ} C$ . What is the heat of reaction at constant volume at $17^{\circ} C$ ?

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Heat of Reaction

Standard XII Chemistry

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What is value of △U (heat change at constant volume) for reversible isothermal evaporation of 90 g water at 100oC. Assuming water vapour behaves as an ideal gas and △Hvap)water=540 cal g−1

The correct option is C 4.49×103 calHeat of system (△H)v=m×Lv =90×540=48600 cal △H=△U+P(VV−VL)=△U+nRT or, 48600=△U+9018×2×373 or, △U=48600−3730=44870 cal

What is value of $△ U$ (heat change at constant volume) for reversible isothermal evaporation of 90 g water at $100^{o} C .$ Assuming water vapour behaves as an ideal gas and $△ H_{v a p})_{w a t e r} = 540 c a l g^{- 1}$

The correct option is C $4.49 \times 10^{3} c a l$
Heat of system $(△ H)_{v} = m \times L_{v}$
$= 90 \times 540 = 48600 c a l$
$△ H = △ U + P (V_{V} - V_{L}) = △ U + n R T$
or, $48600 = △ U + \frac{90}{18} \times 2 \times 373$
or, $△ U = 48600 - 3730 = 44870 c a l$