Q#19.6
A gas undergoes two processes. In the first, the volume remains constant at 0.200 $m^3$ and the pressure increases from 2.00 $\times 10^5$ Pa to 5.00 $\times 10^5$ Pa. The second process is a compression to a volume of 0.120 $m^3$ at a constant pressure of 5.00 $\times 10^5$ Pa.
(a) In a pV-diagram, show both processes.
(b) Find the total work done by the gas during both processes.
Answer:
(a) The pV-diagram is sketched in Figure 19.6.
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| Fig.19.6 |
(b) Calculate W for each process, using the expression for W that applies to the specific type of process. 1 → 2 , ΔV = 0, so W = 0.
For 2 → 3, p is constant; so
W = pΔV = $(5.00 \times 10^5 \ Pa)(0.120 \ m^3 - 0.200 \ m^2)= -4.00 \times 10^4 \ J$
(W is negative since the volume decreases in the process.)
$W_{total} = W_{1→2} + W_{2→3}=-4.00 \times 10^4$ J
The volume decreases so the total work done is negative.
Q#19.7
Work Done in a Cyclic Process.
(a) In Fig. 19.7a, consider the closed loop 1 → 3 → 2 → 4 → 1. This is a cyclic process in which the initial and final states are the same. Find the total work done by the system in this cyclic process, and show that it is equal to the area enclosed by the loop.
(b) How is the work done for the process in part (a) related to the work done if the loop is traversed in the opposite direction, 1 → 4 → 2 → 3 → 1, Explain.
Answer:
Calculate W for each step using the appropriate expression for each type of process.
When p is constant, W = p$\Delta V$. When $\Delta V = 0$, W = 0.
(a) $W_{13}=p_1(V_2-V_1)$, $W_{32}=0$, $W_{24}=p_2(V_1-V_2)$ and $W_{41}=0$.
The total work done by the system is
$W_{13}+W_{32}+W_{24}+W_{41}=(p_1-p_2)(V_2-V_1)$
which is the area in the pV plane enclosed by the loop.
(b) For the process in reverse, the pressures are the same, but the volume changes are all the negatives of those found in part (a), so the total work is negative of the work found in part (a).
When $\Delta V > 0$, W > 0 and when $\Delta V < 0$, W < 0.
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