Question

A gas is in a container with an initial volume of 2.50L, initial pressure of 1.34 atm, and initial temperature of 308 K. What would be the pressure if the gas was heated to 373 K and the volume decreased to 1.90 L? *Hint use the Combined Gas Law.

Answer 1

Answer: The pressure is 2.14 atm if the gas was heated to 373 K and the volume decreased to 1.90 L.

Explanation:

Given: $V_{1} = 2.50 L$,      $P_{1} = 1.34 atm$,       $T_{1} = 308 K$

$P_{2} = ?$ ,            $V_{2} = 1.90 L$,          $T_{1} = 373 K$

Formula used to calculate the final pressure is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1.34 atm \times 2.50 L}{308 K} = \frac{P_{2} \times 1.90 L}{373 K}\\P_{2} = \frac{1.34 atm \times 2.50 L \times 373}{308 K \times 1.90 L}\\= 2.14 atm$

Thus, we can conclude that the pressure is 2.14 atm if the gas was heated to 373 K and the volume decreased to 1.90 L.