Answer:
The gas will occupy a volume of 4000L
Explanation:
V1 = 2000L
T1 = 100K
P1 = 100kPa = 100*10³Pa
T2 = 400k
P2 = 200kPa = 200*10³kPa
V2 = ?
To solve this question, we need to use combined gas equation
[(P₁ * V₁) / T₁] = [(P₂ * V₂) / T₂]
V₂ = (P₁ * V₁ * T₂) / (P₂ * T₁)
V₂ = (100*10³ * 2000 * 400) / (200*10³ * 100)
V₂ = 8.0*10¹⁰ / 2.0*10⁷
V₂ = 4000L
The new volume of the gas is 4000L
Answer:
The volume that the gas will occupy when the temperature is increased to 400.0 K and the pressure is increased to 200. kPa is 4000. L
Explanation:
Here we are required to utilize the combined gas equation as follows;
[tex]\frac{P_{1}\times V_{1}}{T_{1}} = \frac{P_{2}\times V_{2}}{T_{2}}[/tex]
Where:
P₁ = Initial pressure of the gas = 100.0 kPa
V₁ = Initial volume of the gas = 2000. L
T₁ = Initial temperature of the gas = 100.0 K
P₂ = Final pressure of the gas = 200.0 kPa
V₂ = Final volume of the gas = Required
T₂ = Final temperature of the gas = 400.0 K
Making V₂ the formula subject of the combined gas equation, we have;
[tex]V_{2}}{} = \frac{P_{1}\times V_{1}\times T_{2} }{T_{1}\times P_{2}}[/tex]
Therefore, by plugging the values, we have;
[tex]V_{2}}{} = \frac{100.0\times 2000\times 400.0 }{100.0 \times 200.0 } = 4000. \, L[/tex]
The volume that the gas will occupy when the temperature is increased to 400.0 K and the pressure is increased to 200. kPa = 4000. L.
