$18.0$ g of water completely vapourises at $100^{o} C$ and 1 bar pressure and the enthalpy change in the process is $40.79 k j m o l^{- 1}$ . What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalpy of vapourisation for water?

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Solution

Amount of heat required to vapourise one mole of a liquid at constant temperature and standard pressure is called its standard enthalpy of vapourization.

One mole of water weighs $18$ g and enthalpy change is $40.79 k j / m o l$ . Since enthalpy of vapourization is given for 1 mole, its standard enthalpy of vapourization for water is same i.e, $40.79 k j / m o l$ .

For 2 mole of water, enthalpy of vapourization will be $2 \times 40.79 = 81.58 k j .$

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18.0 g of water completely vaporises at $100^{\circ} C$ and 1 bar pressure and the enthalpy change in the process is 40.79 kJ $m o l^{- 1}$ . What will be the enthalpy change for vaporising two moles of water under the same conditions? What is the standard enthalpy of vaporisation for water?

Q. 18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ mol –1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalphy of vapourisation for water?

Q. Standard enthalpy of vapourisation

Δ_{v a p} H^{0}

for water at

100^{0} C

40.66 k J . m o l^{- 1}

. The internal energy change of vapourisation of water at

100^{0} C

(i n k J . m o l^{- 1})

is:

Assuming that water vapour is an ideal gas, the internal energy change $(Δ U)$ when 1 mol of water is vapourised at 1 bar pressure and 100 $^{0}$ C will be:

[Given molar enthalpy of vapourisation of water at 1bar and 373 K=41 kJ mol $^{- 1}$ ]

Q. Standard enthalpy of vapourisation

Δ_{v a p} H^{(0)}

for water at

100

is 40.66

k J {m o l}^{- 1}

. The internal energy of vapourisation of water at

100

(in

k J {m o l}^{- 1}

) is:

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18.0 g of water completely vapourises at 100oC and 1 bar pressure and the enthalpy change in the process is 40.79 kj mol−1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalpy of vapourisation for water?

$18.0$ g of water completely vapourises at $100^{o} C$ and 1 bar pressure and the enthalpy change in the process is $40.79 k j m o l^{- 1}$ . What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalpy of vapourisation for water?