Question

A gas is contained in a 5.0–liter container at a pressure of 100 kPa. What size container would be required to contain the gas at a pressure of 50 kPa, assuming the temperature remains constant?

Answer 1

P1* V1 = P2 * V2

100* 5 = 50 * V2

500 = 50 V2

V2 = 500 / 50

V2 = 10.0 L

hope this helps!

Answer 2

Answer : The volume of gas is, 10 L

Explanation:

According top the Boyle's Law, the pressure of a gas is inversely proportional to the volume of the gas.

`P\propto (1)/(V)`

or,

`(P_1)/(P_2)=(V_2)/(V_1)`

where,

`V_1` = initial volume of gas = 5 L

`V_2` = final volume of gas = ?

`P_1` = initial pressure of gas = 100 kPa

`P_2` = final pressure of gas = 50 kPa

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

`(100kPa)/(50kPa)=(V_2)/(5L)`

`V_2=10L`

Therefore, the volume of gas is, 10 L