Expansion of Work 8.3 Self Test


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Kawsar Nasir 3k
Posts: 21
Joined: Fri Sep 25, 2015 3:00 am

Expansion of Work 8.3 Self Test

Postby Kawsar Nasir 3k » Wed Jan 06, 2016 10:28 pm

I don't understand how the density of water and ice are calculated to give e the final work, how is water converted to ice? How is the work calculated afterwards?
"Find the work of 100g of H2O when it freezes and pushes back the metal wall of a pipe that exerts an opposing pressure of 1070 atm? Density of water and ice at 0.0 degrees is 1.00 gcm-3 and .92gcm-3"
Thanks

Nathan Danielsen 1G
Posts: 17
Joined: Fri Sep 25, 2015 3:00 am

Re: Expansion of Work 8.3 Self Test

Postby Nathan Danielsen 1G » Thu Jan 07, 2016 10:19 am

I think you can do it like this.

To find the work here you need to use the equation W=-P * deltaV. In the problem you are given the pressure, 1070 atm, so you just need to find the volume. The change in volume will be the final volume, minus the initial volume, so you will need to find the volume of the water when it is liquid, and when it is solid. You know you have 100 g of water, and the phase change from liquid to solid does not change the amount of water present. Therefore to find the volume in cm^3 of liquid water and ice, you must divide 100 by their respective densities. For liquid water, 100 g / (1.00 g/cm^3)= 100, which means that the 100 grams of liquid water take up 100 cm^3. For ice, 100 g / (.92 g/cm^3)=108.70, meaning that the 100 grams of ice take up 108.7 cm^3. To convert this to liters, use the conversion that 1 L = 1000 cm^3. After this you end up with .1 liters of liquid water initially, and .1087 liters of ice at the end. Subtract the initial value from the final value to get .0087 liters. Then multiply that by -1070 atm to get -116 L*atm. Finally convert L*atm to J, by multiplying -116 L*atm by 101.325 J/L*atm. You end up with 12,000 J (adjusted for sig figs).


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