Consider a horizontal cylinder-and-piston device similar to the one shown in Fig
ID: 939738 • Letter: C
Question
Explanation / Answer
a) the temperature is constant so the system is isothermal
b) There is no change in temperature so the change in internal energy = 0
Also Heat = work done
the work done will be positive as the gas is being compressed
Work done = work initial - workfinal = nRT ln (Vf / Vi)
Now the external pressure = 2X internal pressure
And volume is inversely proportional to pressure
So the final volume = half of intial volume
n = moles = 0.5
R = 8.314
T = 300K
c) work done = 0.5 X 8.314 X 300 ln (ln(1/2) = 864.24 Joules
Heat = -864.24 Joules
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