Question

9)

I have an unknown volume of gas held at a temperature of

115 Kin á container with a pressure of 60 atm. If by

increasing the temperature to 225 K and decreasing the

pressure to 30 atm causes the volume of the gas to be 29

liters, how many liters of gas did I start with?

Answer 1

Answer : The volume of gas will be, 113.5 L

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 = 60 atm

$P_2$ = final pressure of gas = 30 atm

$V_1$ = initial volume of gas = 29 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = 115 K

$T_2$ = final temperature of gas = 225 K

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

$\frac{60atm\times 29L}{115K}=\frac{30atm\times V_2}{225K}$

$V_2=113.5L$

Therefore, the volume of gas will be, 113.5 L