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Work done on/by hydrogen gas

  • Thread starter Shayna
  • Start date
  • #1
13
0

Homework Statement


The figure shows a thermodynamic process followed by 1.80×10−2 mol of hydrogen.
knight_Figure_17_65.jpg


How much work is done on the gas?
By how much does the thermal energy of the gas change?
How much heat energy is transferred to the gas?


The attempt at a solution
I thought the work done to gas should be negative the area underneath the curve, which is
W= [(300-100)*0.001L * (4-1)atm] = (3 atm * 0.2L)/2 = 0.3 L*atm = 0.3*101.3 J = 30.39 J
apparently that is neither here nor there

Also, the thermal energy of the gas change would be the same as the amount of work done on the gas plus heat energy transferred to the gas, is that correct?

Pleeeeeease hep, I am so confused.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Hi Shayna! :smile:

Hint: work done = force times change in position = pressure times change in volume. :wink:
 
  • #3
13
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enn, well, I calculated the area wrong, but how do I find out about the heat transfered?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hi Shayna! :smile:
How much work is done on the gas?
By how much does the thermal energy of the gas change?
How much heat energy is transferred to the gas?

Also, the thermal energy of the gas change would be the same as the amount of work done on the gas plus heat energy transferred to the gas, is that correct?
enn, well, I calculated the area wrong, but how do I find out about the heat transfered?
(I think thermal energy and heat energy are the same … wikipedia certainly seems to think so: http://en.wikipedia.org/wiki/Heat_energy)

Yes, from the work-energy theorem, work done = energy transferred. :smile:

(that's the only reason anyone is interested in calculating work :wink:)
 
  • #5
13
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I got the work done by calculating area under the curve which is -50.7J
I also calculated the temperature change delta T= T1-T2 = nR/(p1V1-p2V2)= -67.75K
I also tried specific heat, which q= Cq * n* delta T = 28.82 J/mol*K * 1.8*10^-2 mol * 67.2585 K = -35.1 J
But that wasn't the answer
I don't know where to go from here
Thanks so much for your help
 
  • #6


I got the work done by calculating area under the curve which is -50.7J
I also calculated the temperature change delta T= T1-T2 = nR/(p1V1-p2V2)= -67.75K
I also tried specific heat, which q= Cq * n* delta T = 28.82 J/mol*K * 1.8*10^-2 mol * 67.2585 K = -35.1 J
But that wasn't the answer
I don't know where to go from here
Thanks so much for your help
I'm stuck on this one myself but one thing I know is that the specific heat is the piece of knowledge missing here since the values for the specific heat one can look up are either c_p(for constant pressure) or c_v (for constant volume) and this problem has neither constant volume nor pressure.
 
  • #7


In just talked to someone about this apparently you use C_v to find the change in thermal energy. So it is just a matter of finding the two temperatures T=PV/nR then the change in temperature T_1-T_2. Then applying (delta)E_th=n*C_v*(delta)T
 

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