Question 188: Thermodynamics & KTG

Question

The given figure represents two isobaric processes for the same mass of an ideal gas, then
![](https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsege4i4/aa4e3a85-6a8b-4bf3-9893-7b3be92c9ee2/c08811b0-c72e-11ee-8cb5-7deab4010c9e/file-6y3zli1lsege4i5.png?format=png)

Accepted Answer

The two isobaric processes depicted in the figure show changes in the volume of an ideal gas with temperature under constant pressure conditions.
Isobaric processes follow the equation $PV = nRT$, where P is pressure, V is volume, n is the number of moles of the gas, R is the ideal gas constant, and T is temperature.
When we rearrange the equation in terms of V (Volume), we get $V = \left(\frac{nR}{P}\right)T$. The slope of the volume-temperature graph in such a scenario is given by $\frac{nR}{P}$. This implies that the slope is inversely proportional to the pressure (i.e., slope $\propto \frac{1}{P}$).
From this relationship, if one process has a higher slope compared to another, its corresponding pressure must be lower. Observing the given figure and applying this understanding, if Process 2 has a greater slope than Process 1, it indicates that $P_2 < P_1$.

Answer

$P_2>P_1$

Answer

$P_1>P_2$

Answer

$P_1=P_2$

Answer

$P_2 \geq P_1$

Thermodynamics & KTG

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

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