Question

A gas of certain mass occupies volume of 1.2 L at 37°C and 3 atm. At what temperature will the volume and pressure of this gas become one-third of their initial values?​

Answer 1

Answer: At a temperature of 34.44 K the volume and pressure of this gas become one-third of their initial values.

Explanation:

Given : $V_{1}$ = 1.2 L,        $T_{1} = 37^{o}C = (37 + 273) K = 310 K$

$P_{1}$ = 3 atm,                  $V_{2} = \frac{1.2}{3} = 0.4 L$ ,      $P_{2} = \frac{3}{3} = 1 atm$

Formula used to calculate the final temperature 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.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{3 atm \times 1.2 L}{310 K} = \frac{1 atm \times 0.4 L}{T_{2}}\\T_{2} = \frac{1 atm \times 0.4 L \times 310 K}{3 atm \times 1.2 L}\\= \frac{124}{3.6} K\\= 34.44 K$

Thus, we can conclude that the final temperature is 34.44 K.