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Standard X

Physics

Unit of Heat

31. 90 g of w...

Question

31. 90 g of water spilled out from a vessel in the room on the floor. Assuming that water vapour behaving as an ideal gas, calculate the internal energy change when the spilled water undergoes complete evaporation at 100 ^° C (Given the molar enthalpy of vaporization of water at 1 bar and 373 K =41 kJmol¯¹)

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Q. Assuming that water vapour is an ideal gas, the internal energy change

(Δ U)

when

1

mol of water is vaporized at

1

bar pressure and

100^{o}

C (given: molar enthalpy of vaporization of water

41

k J

m o l^{- 1}

1

bar and

373

K and

R = 8.3

m o l^{- 1}

m o l^{- 1}

) will be:

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 $^{- 1}$ ]

Assuming that, water vapour is an ideal gas the internal energy change (ΔU)(ΔU) when 11 mol of water is vaporized at 11 bar pressure and 100o100o C will be:

[given: molar enthalpy of vaporisation of water at

1

bar and

373

= 41

m o l^{- 1}

and

R = 8.3

K^{- 1}

m o l^{- 1}

]

If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourization of 1 mol of water at 1 bar and

100^{0} C

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

. Calculate the internal energy change, when 1 mol of water is vaporized at 1 bar pressure and

100^{0} C

Q. Assuming that water vapour is an ideal gas, the internal energy change (

Δ U

) when

1 m o l

of water is vaporised at

1

bar pressure and

100^{o} C

will be:

(Given: Molar enthalpy of vapourisation of water at

1

bar and

373 K = 41 k J . {m o l}^{- 1}

and

R = 8.3 J {m o l}^{- 1}

K^{- 1}

)

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