# Work done on/by hydrogen gas

## Homework Statement

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

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tiny-tim
Homework Helper
Hi Shayna! Hint: work done = force times change in position = pressure times change in volume. enn, well, I calculated the area wrong, but how do I find out about the heat transfered?

tiny-tim
Homework Helper
Hi Shayna! 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. (that's the only reason anyone is interested in calculating work )

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
I don't know where to go from here
Thanks so much for your help

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