Answer:
The gas will occupy a volume of 1.702 liters.
Explanation:
Let suppose that the gas behaves ideally. The equation of state for ideal gas is:
[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex] (1)
Where:
[tex]P[/tex] - Pressure, measured in kilopascals.
[tex]V[/tex] - Volume, measured in liters.
[tex]n[/tex] - Molar quantity, measured in moles.
[tex]T[/tex] - Temperature, measured in Kelvin.
[tex]R_{u}[/tex] - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.
We can simplify the equation by constructing the following relationship:
[tex]\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}[/tex] (2)
Where:
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressure, measured in kilopascals.
[tex]V_{1}[/tex], [tex]V_{2}[/tex] - Initial and final volume, measured in liters.
[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Initial and final temperature, measured in Kelvin.
If we know that [tex]P_{1} = 121\,kPa[/tex], [tex]P_{2} = 202\,kPa[/tex], [tex]V_{1} = 2.7\,L[/tex], [tex]T_{1} = 288\,K[/tex] and [tex]T_{2} = 303\,K[/tex], the final volume of the gas is:
[tex]V_{2} = \left(\frac{T_{2}}{T_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot V_{1}[/tex]
[tex]V_{2} = 1.702\,L[/tex]
The gas will occupy a volume of 1.702 liters.
