Question

The pressure inside a gas bottle is 4120 kPa when the temperature is 29.0 °C. What will be the pressure inside the bottle, if 34.9 percent of the gas is released and the temperature of the gas drops to 5.5 °C?

Answer 1

Answer: The pressure inside the bottle is 2473 kPa

Step-by-step explanation:

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

The combined gas equation is,

`(P_1V_1)/(n_1RT_1)=(P_2V_2)/(n_2RT_2)`

where,

`P_1` = initial pressure of gas = 4120 kPa

`P_2` = final pressure of gas = ?

`V_1` = initial volume of gas = v

`V_2` = final volume of gas = v

`n_1` = initial moles of gas = n

`n_2` = final moles of gas =
`n-(34.9)/(100)* n=0.651n`

`T_1` = initial temperature of gas =
`29.0^oC=273+29.0=302.0K`

`T_2` = final temperature of gas =
`5.5^oC=273+5.5=278.5K`

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

`(4120* v)/(n* R* 302.0)=(P_2* v)/(0.651n* R* 278.5)`

`P_2=2473kPa`

Thus the pressure inside the bottle, if 34.9 percent of the gas is released and the temperature of the gas drops to 5.5 °C is 2473 kPa