Question

For the thermodynamic cycle shown in Fig. $20-3$, find $(a)$ the net work output of the gas during the cycle and $(b)$ the net heat flow into the gas per cycle. Method 1 (a) From Problem 20.13, the net work done is $1.0 \mathrm{~J}-0.50 \mathrm{~J}=0.5 \mathrm{~J}$. Method 2 The net work done is equal to the area enclosed by the $P-V$ diagram: $$ \text { Work }=\text { Area } A B C D A=\left(2.0 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(2.5 \times 10^{-6} \mathrm{~m}^{3}\right)=0.50 \mathrm{~J} $$ (b) Suppose the cycle starts at point- $A$. The gas returns to this point at the end of the cycle, so there is no difference in the gas at its start and end points. For one complete cycle, $\Delta U$ is therefore zero. We have then, if the first law is applied to a complete cycle, $$ \Delta Q=\Delta U+\Delta W=0+0.50 \mathrm{~J}=0.50 \mathrm{~J}=0.12 \mathrm{cal} $$

          For the thermodynamic cycle shown in Fig. $20-3$, find $(a)$ the net work output of the gas during the cycle and $(b)$ the net heat flow into the gas per cycle.
Method 1
(a) From Problem 20.13, the net work done is $1.0 \mathrm{~J}-0.50 \mathrm{~J}=0.5 \mathrm{~J}$.
Method 2
The net work done is equal to the area enclosed by the $P-V$ diagram:
$$
\text { Work }=\text { Area } A B C D A=\left(2.0 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(2.5 \times 10^{-6} \mathrm{~m}^{3}\right)=0.50 \mathrm{~J}
$$
(b) Suppose the cycle starts at point- $A$. The gas returns to this point at the end of the cycle, so there is no difference in the gas at its start and end points. For one complete cycle, $\Delta U$ is therefore zero. We have then, if the first law is applied to a complete cycle,
$$
\Delta Q=\Delta U+\Delta W=0+0.50 \mathrm{~J}=0.50 \mathrm{~J}=0.12 \mathrm{cal}
$$

Added by Samantha R.

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