Question

Can someone help me answer this heat/thermodynam. question? the answers are supposed to be a. -1.69*10^5J, b. 2.08*10^6 J, and c.1.70 m^3: The latent heat of vaporization for water is 539 kcal/kg at 100
C. One kilogram of steam occupies 1.67 m3 at
100
C and 1 atm.
a.) How much work is done in converting 1.00 kg of water to steam at 100
C and 1 atm?
b.) What is the change in internal energy when 1.00 kg of water is converted to steam at 100
C and 1 atm?
c.) Use the ideal gas law to calculate the volume of 1.00 kg of steam at 100
C and 1 atm.

Answer 1

c) pV = n R T so V = n R T/ p

1 mole (18 gm) of H2O gas would occupy 22.4 L at STP = 0o C, 1 atm

Work with the ratios, as R and p are the same

Here V/Vstd = (1000/18)(373/273)(22.4 L) =1700 L = 1.70 m^3.

This is a very large change in volume from 1L liquid H2O

This answer is useful for a and b as dU = Q - pdV

Answer 2

thank you!!!