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

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,

`(P_1V_1)/(T_1)=(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.

`(319mmHg* 0.558L)/(115K)=(215mmHg* V_2)/(387K)`

`V_2=2.79L`

Therefore, the volume of gas will be, 2.79 L

Answer 2

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.

`(P1V1)/(T1) = (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