#### buttterfly41

1. A gas is compressed from 10.00 L to 2.00 L at a constant pressure of 0.800 atm. In the process, 400 J of energy leaves the gas by heat.
(a) What is the work done by the gas?
(b) What is the change in its internal energy?

so for part a, i have tried:
P*deltaV=W
8.106E4 * 8 = 6.48E5 for the answer to part a, it is wrong and i dont know how else to approch this. I put the pressure in Pa, but maybe i should have left it in atm? i dont know, any help would be wonderful. And for part b, i know that Q-W=U, so i should be able to get b, once i figure out part a.

2. Gas in a container is at a pressure of 1.30 atm and a volume of 3.00 m3.
(a) What is the work done by the gas if it expands at constant pressure to twice its initial volume?
(b) If it is compressed at constant pressure to one quarter of its initial volume?

So far for part a, i have tried:
W=P*delta V
1.32E5 * 3000L = 3.96E8 J, but this isnt right...
I also tired:
W=Pln(vf/vi)
1.32E5* ln(6000/3000) = 9.15E4 J... again wrong
not sure what else to try, and again, part b will be much the same once i understand part a

lastly,
3. A sample of helium behaves as an ideal gas as energy is added by heat at constant pressure from 273 K to 393 K. If the gas does 15.0 J of work, what is the mass of helium present?

so i have tried:
Q=mc*deltaT
15=m*5.193*120
m=.0241moles *4g/mol = .0482g... but wrong

Any and all help/suggestiong would be much appreciated thankyou.

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#### Hootenanny

Staff Emeritus
Gold Member
Question 1.(a) seems a logical place to begin. Think about question (a) again; is the gas doing work or are you doing work on the gas?

#### Andrew Mason

Homework Helper
2. Gas in a container is at a pressure of 1.30 atm and a volume of 3.00 m3.
(a) What is the work done by the gas if it expands at constant pressure to twice its initial volume?
(b) If it is compressed at constant pressure to one quarter of its initial volume?

So far for part a, i have tried:
W=P*delta V
1.32E5 * 3000L = 3.96E8 J, but this isnt right...
Your method is correct (dW = PdV so $W = \int dW = P\int dV = P(V_f - V_i)$). Check your units. If you are using MKS you cannot use litres.
3. A sample of helium behaves as an ideal gas as energy is added by heat at constant pressure from 273 K to 393 K. If the gas does 15.0 J of work, what is the mass of helium present?

so i have tried:
Q=mc*deltaT
15=m*5.193*120
m=.0241moles *4g/mol = .0482g... but wrong
Since you know that PV = nRT, create an expression for W in terms of temperatures and determine n from that.

I am not sure of your thinking using the heat approach.

Since $dQ = nC_pdT = dU + dW = nC_vdT + PdV$ you can work out n. What is the difference in specific heats (Cp - Cv)?

AM

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#### buttterfly41

thanks to both of you, i was confused, but i get it now! THANKS1!!