ideal gas?

(a) First of all, assume that the process is done under 1 atmospheric pressure, i.e. 1.01 × 105 Pa (pascal)
By using the ideal gas equation PV = nRT, we have:
V = nRT/P
= 1 × 8.314 × 350 / (1.01 × 105)
= 0.0288 m3
Therefore, if it is compressed to half of its original volume, then △V = -0.288/2 = -0.144 m3. (Negative because this is a decrease in volume)
So by W = P × △V, where W is the work done, the work done by the gas is:
1.01 × 105 × (-0.144) = -1455 J
in which the negative sign means that work is done ON the gas, i.e. the gas has gained energy in the process.
(b) According to the equation:
△U = Q + W
where △U = change in internal energy
Q = heat supplied
W = work done
In the second part of the process, it is done at fixed volume and so △V, and hence work done = 0.
So △U for the second part of the process = 400 J.
Finally, the total △U for the whole process, in addition to that in the first part, is 1455 + 400 = 1855 J.

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