Question

The given figure represents two isobaric processes for the same mass of an ideal gas, then

• A. $P_2 \geq P_1$
• B. $P_2 > P_1$
• C. $P_1 = P_2$
• D. $P_1 > P_2$

Answer 1

Answer: (D)

$P_1 > P_2$

Answer 2

From the ideal gas law:

$PV = nRT$

Rearranging for volume:

$V = \left( \frac{nR}{P} \right) T$

The slope of the line in the $V-T$ graph for an isobaric process is proportional to $\frac{1}{P}$. Therefore, we have:

$\text{Slope} \propto \frac{1}{P}$

Comparing slopes: $(\text{Slope})_2 > (\text{Slope})_1 \quad \implies \quad P_2 < P_1$