Work done during expansion of a gas

[tanzerino]

[PLAIN]http://img101.imageshack.us/img101/8176/24851738.png

it is definetaly negative as Eint=Q+W
and Q is smaller than the area under curve taking care of the scaling of pressure in atm.
but what i get is -1.5x10^6

 

[tanzerino]

Einternal=Q+W
Q he said is given so +ve and work from the graph is -ve
calculated work was much greateer that's why i think the answer is -ve then wen i tried the numbers i get -1.5 which isn't an answer

 

[tanzerino]

any help?

 

[Andrew Mason]

You should show your work. But I get the same answer as you do.

$$\Delta Q = \Delta U + W$$

where W is the work done by the gas.

That work is the area under the graph. There appears to be an increase in volume of 5 m^3 and a drop in pressure from 7 to 3 atm. So the work done is (7+3)*5/2 = 25 atm-m^3 of work = 2.525e6 Joules.

So:

$$\Delta U = \Delta Q - W = 1.02e6 - 2.525e6 = -1.5e6 J$$

AM

 

[tanzerino]

thanks for your help.i will check for any mistakes with the teacher this is cengage and the second mistake already.shame!