Question

The pressure exerted by 10 litres of gas at 272 K is 1.6 atm. What pressure would be exerted, in atm, by the gas at 409 K, provided that the volume remains constant?

• A. 2
• B. 2.4
• C. 2.8
• D. 3.2

Answer 1

The correct option is B 2.4

This question deals with changes in pressure and temperature with the volume remaining constant.

This is dealt by usine Gay Lussac's law, which states that volume remaining constant, the pressure exerted by a gas is directly proportional to its temperature.

i.e $P\alpha T$ or $\frac{P}{T}=k$ where $k$ is a constant.

Thus, $\frac{P_1}{T_1}=\frac{P_2}{T_2}$ where $P_{1}and{T}_1$ are the initial pressure and temperature, where as $P_{2}and{T}_2$ are the final ones.

According to the question, $P_1$= 1.6 atm, $T_1$= 272K, $T_2$= 409 K, $P_2$ = ?

Thus, $\frac{1.6}{272}=\frac{P_2}{409}$

Hence, solving the equation, we get $P_2$= 2.4 atm.