Question

A gas is contained in a thick-walled balloon. When the pressure changes from 319 mm Hg to 215 mm Hg, th volume changes from 0.558 L to L and the temperature changes from 115 K to 387 K.

Answer 1

In this question we need to find the new volume of the gas. Since we have been given the pressure and temperature change, we can used to combined gas law equation.
$\frac{P1V1}{T1} = \frac{P2V2}{T2}$
the parameters for 1st instance are given on the left side and parameters for the second instance are given on the right side of the equation
(319 mmHg x 0.558 L)/ 115 K = (215 mmHg x V)/387 K
V = 2.79 L

Answer 2

Answer : The volume of gas will be, 2.79 L

Solution :

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 = 319 mmHg

$P_2$ = final pressure of gas = 215 mmHg

$V_1$ = initial volume of gas = 0.558 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = 115 K

$T_2$ = final temperature of gas = 387 K

Now put all the given values in the above equation, we get the final volume of gas.

$\frac{319mmHg\times 0.558L}{115K}=\frac{215mmHg\times V_2}{387K}$

$V_2=2.79L$

Therefore, the volume of gas will be, 2.79 L