A container of 1.00 mole of an ideal gas is compressed from i to f along either

Question

A container of 1.00 mole of an ideal gas is compressed from i to f along either path 1 or path 2, shown on the p-V graph. Assume 3 significant figures for all values on the graph. Circle the correct term within each set of parentheses: Path 1 is first isobaric, isochoric, isothermal) and then (isobaric, isochoric isothermal). The path 2 curve from i to f represents an isothermal process. Determine the temperature along this curve. For each paths, 1 and 2, determine (i) the work W done by the gas, (ii) the internal energy change delta U_int for the gas, and (iii) the heat Q added to the gas.

Explanation / Answer

b)
Select any point in the isothermal process
Consider the point i
P = 1 x 105 Pa, V = 0.04 m3
Using the ideal gas law
PV = nRT
T = PV/nR
= [(1 x 105) x (0.04)] / [1 x 8.314]
= 481.12 K

c)

Path 2
Work done = integral of PdV
From ideal gas law, P = nRT/V
pdV = nRT [dV/V]
integrating, W = nRT ln [Vf/Vi]
= 1 x 8.314 x 481.12 x ln [0.01/0.04]
= - 5545.18 J (negative sign indicates the compression of gas)

Change in internal energy, dU = 0 for isothermal process since dT = 0

Take Q be the heat added
Q = dU + W
dU = 0
Q = W
= - 5545.18 J (Heat is taken out of the gas)

Path 1.
Work done
Work done for isobaric process = PdV
= 1 x 105 x (0.01 - 0.04)
= - 3000 J
Work done for isochoric process = 0 since dV = 0
Total work done =  -3000 + 0
= - 3000 J

Change in internal energy
This is same as that of path 2 since change in internal energy is a state function. It doesn't depend on the path taken to reach from i to f
Change in internal energy = 0

Heat added
Heat added. Q = dU + W
= -3000 J

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