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(B) For an ideal gas, the equation of state is $PV = nRT$, which can be written as $P = (nR/V)T$.$1$. Process $A$: The line is vertical, meaning $T$ i

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$B -$ Isobaric process

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(B) For an ideal gas, the equation of state is $PV = nRT$, which can be written as $P = (nR/V)T$.$1$. Process $A$: The line is vertical, meaning $T$ is constant. This is an isothermal process.$2$. Process $B$: The line is horizontal, meaning $P$ is constant. This is an isobaric process.$3$. Process $C$: The line passes through the origin, so $P \propto T$. Since $P = (nR/V)T$, this implies $V$ is constant. This is an isochoric process.$4$. Process $D$: The slope of the $P-T$ graph is $dP/dT$. For an adiabatic process, $PV^{\gamma} = 	ext{constant}$. Substituting $V = nRT/P$, we get $P(T/P)^{\gamma} = 	ext{constant}$, which simplifies to $P^{1-\gamma} T^{\gamma} = 	ext{constant}$. Differentiating this, we find that the slope for an adiabatic process is steeper than the isochoric process in a $P-T$ diagram.Comparing the options, $B$ represents an isobaric process.

The following figure shows a $P-T$ graph for four processes $A, B, C$ and $D$. Select the correct alternative.

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