Question

Some hydrogen gas is enclosed within a chamber being held at 200^\ { C} with a volume of 0.025 \rm m^3. The chamber is fitted with a movable piston. Initially, the pressure in the gas is 1.50 \times 10^6 \; \rm Pa (14.8 \rm atm). The piston is slowly extracted until the pressure in the gas falls to 0.950 \times 10^6 \; \rm Pa. What is the final volume V_2 of the container? Assume that no gas escapes and that the temperature remains at 200^ { C}.

Answer 1

The final volume is
`0.039 m^3`

Step-by-step explanation:

Data:

Initial temperature:
`T1=200C`

Final temperature:
`T2=200C`

Initial pressure:
`P1=1.50 *10^6 Pa`

Final pressure:
`P2=0.950 *10^6 Pa`

Initial volume:
`V1=0.025m^(3)`

Final volume:
`V2=?`

Assuming hydrogen gas as a perfect gas it satisfies the perfect gas equation:

`(PV)/(T)=nR` (1)

With P the pressure, V the volume, T the temperature, R the perfect gas constant and n the number of moles. If no gas escapes the number of moles of the gas remain constant so the right side of equation (1) is a constant, that allows to equate:

`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`

Subscript 2 referring to final state and 1 to initial state.

solving for V2:

`V_(2)=(P_(1)V_(1)T_(2))/(T_(1)P_(2))=((1.50 *10^6)(0.025)(200))/((200)(0.950 *10^6))`

`V_(2)=0.039 m^3`