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

Answer:

The final volume is $0.039 m^3$

Explanation:

Data:

Initial temperature: $T1=200C$

Final temperature: $T2=200C$

Initial pressure: $P1=1.50 \times10^6 Pa$

Final pressure: $P2=0.950 \times10^6 Pa$

Initial volume: $V1=0.025m^{3}$

Final volume: $V2=?$

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

$\frac{PV}{T}=nR$ (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:

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

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

solving for V2:

$V_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}=\frac{(1.50 \times10^6)(0.025)(200)}{(200)(0.950 \times10^6)}$

$V_{2}=0.039 m^3$