Question

The volume-temperature graphs of a given mass of an ideal gas at constant pressures are shown below. What is the correct order of pressures?

• A. $P_1>P_3>P_2$
• B. $P_1>P_2>P_3$
• C. $P_2>P_3>P_1$
• D. $P_2>P_1>P_3$

Answer 1

The correct option is B $P_1>P_2>P_3$

The combined gas law states that the pressure of a gas is inversely related to the volume and directly related to the temperature. So as volume increases pressure decreases. So the correct order of pressure is $P_1>P_2>P_3$.
Slope $=m=\frac{V}{T}\propto \frac{1}{P}$
From the above relation higher the slope of the given curve lower is the respective pressure.
Hence the correct answer is option B.