Question

Some hydrogen gas is enclosed within a chamber being held at 200∘C with a volume of 0.0250 m3. The chamber is fitted with a movable piston. Initially, the pressure in the gas is 1.50×106Pa (14.8 atm). The piston is slowly extracted until the pressure in the gas falls to 0.950×106Pa. What is the final volume V2 of the container

Answer 1

Answer:

0.04 m³.

Explanation:

• We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

• If n and T are constant, and have different values of P and V:

(P₁V₁) = (P₂V₂).

V₁ = 0.025 m³, P₁ = 1.5 x 10⁶ Pa,

V₂ = ??? m³, ​P₂ = 0.95 x 10⁶ Pa.

∴ V₂ = (P₁V₁)/(P₂) = (0.025 m³)(1.5 x 10⁶ Pa)/(0.95 x 10⁶ Pa) = 0.04 m³.

Answer 2

Answer: The final volume of the container is $0.040m^3$

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.  

The equation given by this law is:

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are initial pressure and volume.

$P_2\text{ and }V_2$ are final pressure and volume.

We are given:

$P_1=1.50\times 10^6Pa\\V_1=0.0250m^3\\P_2=0.950\times 10^6\\V_2=?m^3$

Putting values in above equation, we get:

$1.50\times 10^6Pa\times 0.0250m^3=0.950\times 10^6Pa\times V_2\\\\V_2=\frac{1.50\times 10^6\times 0.0250}{0.950\times 10^6}=0.040m^3$

Hence, the final volume of container is $0.040m^3$