Question

A) Determine the work done by 4.00 kg of water when it is all boiled to steam at 100 ∘C. Assume a constant pressure of 1.00 atm.

B) Determine the change in internal energy of 4.00 kg of water when it is all boiled to steam at 100 ∘C. Assume a constant pressure of 1.00 atm.

Answer 1

Solution :-

Given data,

Mass of water= 4 kg

Temperature = 100°c

Pressure = 1.0 atm

Density of water = 1000 kg/m^3

1 mol of water=18 grams

4kg of water=222.222 mols

volume of water=0.004 m^3

assuming steam to be ideal gas,

we can use ideal gas equation to find the volume of the generated steam.

volume=(number of moles×gas constant×temperature)/(pressure)

=(222.222*8.314*)(273+100)/101325 =6.80125 m^3

so work done by water=pressure*(final volume-initial volume)=101325*(6.80125-0.004)=688731.3563 J

heat required to boil m kg of water at 100 degree celcius to steam=m*latent heat of vaporization of water

=m*2264.76 kJ

=4*2264.76*1000 J=9059.04*10^3 J

part A:
work done by the system=688731.3563 J

part B:

change in internal energy=heat supplied - work done by the system

=8370308.644 J