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(D) From the graph, at point $x$: $T_1 = 300 \ K, V_1 = 20 \ L$. At point $z$: $T_2 = 600 \ K, V_2 = 40 \ L$. According to the ideal gas law, $pV = nR

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isobaric

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(D) From the graph, at point $x$: $T_1 = 300 \ K, V_1 = 20 \ L$. At point $z$: $T_2 = 600 \ K, V_2 = 40 \ L$. According to the ideal gas law, $pV = nRT$, which implies $\frac{pV}{T} = nR$. Since $n$ and $R$ are constants, $\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}$. Substituting the values: $\frac{p_1 	imes 20}{300} = \frac{p_2 	imes 40}{600}$. $\frac{p_1}{15} = \frac{p_2}{15}$, which gives $p_1 = p_2$. Since the pressure remains constant during the process from $z$ to $x$, it is an isobaric process.

The following graph shows a system containing $1 \ mole$ of gas undergoing various steps. When it moves from $z$ to $x$, the type of process is:

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