Question 375.
Initially 0.800 mol of an ideal gas in a container occupies a volume of 3.10 l at a pressure of 3.90 atm with an internal energy U1 = 364.8 J. The gas is cooled at a constant volume until its pressure is 2.50 atm. Then it is allowed to expand at constant pressure until its volume is 6.20 l. The final internal energy is U2 = 467.7 J. All processes are quasi static. Draw this process on a PV diagram. What is the work done by the gas? What is the heat absorbed by the gas?
Answer 375.
The processes are shown on PV diagram as under.
Work done by the gas in isochoric process = 0
Work done bt the gas in isobaric process = PdV
= (2.50) * (6.20 - 3.10) lit-atm
= 7.75 lit-atm
= 785.26875 joule ... [Refer to the link - http://www.convertunits.com/from/L+*+atm…
Heat absorbed by the gas = Increase in internal energy + work done by the gas
= 467.7 - 364.8 + 785.3 joule
= 888.2 joule
= (888.2) * (0.239) calories
= 212.3 calories.
[No. of moles of the gas is not needed in any of the above calculations.]
Link to YA!
Initially 0.800 mol of an ideal gas in a container occupies a volume of 3.10 l at a pressure of 3.90 atm with an internal energy U1 = 364.8 J. The gas is cooled at a constant volume until its pressure is 2.50 atm. Then it is allowed to expand at constant pressure until its volume is 6.20 l. The final internal energy is U2 = 467.7 J. All processes are quasi static. Draw this process on a PV diagram. What is the work done by the gas? What is the heat absorbed by the gas?
Answer 375.
The processes are shown on PV diagram as under.
Work done bt the gas in isobaric process = PdV
= (2.50) * (6.20 - 3.10) lit-atm
= 7.75 lit-atm
= 785.26875 joule ... [Refer to the link - http://www.convertunits.com/from/L+*+atm…
Heat absorbed by the gas = Increase in internal energy + work done by the gas
= 467.7 - 364.8 + 785.3 joule
= 888.2 joule
= (888.2) * (0.239) calories
= 212.3 calories.
[No. of moles of the gas is not needed in any of the above calculations.]
Link to YA!
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