Question

A 6.87-L sample of gas has a pressure of 0.732 atm and a temperature of 95 °C. The sample is allowed to expand to a volume of 9.22 L and is cooled to 44 °C. Calculate the new pressure of the gas, assuming that no gas escaped during the experiment.

Answer 1

Answer: The new pressure of the gas, assuming that no gas escaped during the experiment is 0.470 atm

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 0.732 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 6.87 L

$V_2$ = final volume of gas = 9.22 L

$T_1$ = initial temperature of gas = $95^oC=273+95=368K$

$T_2$ = final temperature of gas = $44^oC=273+44=317K$

Now put all the given values in the above equation, we get:

$\frac{0.732\times 6.87}{368}=\frac{P_2\times 9.22}{317}$

$P_2=0.470atm$

Thus the new pressure of the gas, assuming that no gas escaped during the experiment is 0.470 atm