Answer:
The new volume of the gas is 0.332 liters.
Explanation:
Let suppose that argon behaves ideally, the equation of state of the ideal gas is:
[tex]P\cdot V = n \cdot R_{u}\cdot T[/tex]
Where:
[tex]P[/tex] - Pressure, measured in torr.
[tex]V[/tex] - Volume, measured in liters.
[tex]n[/tex] - Molar quantity, measured in moles.
[tex]T[/tex] - Temperature, measured in kelvins.
[tex]R_{u}[/tex] - Ideal gas constant, measured in [tex]\frac{torr\cdot L}{mol \cdot K}[/tex].
Since gas sample is a closed system experimenting a heating process, that is, molar quantity remains constant, the following relationship is derived from the equation of state described above:
[tex]\frac{P_{o}\cdot V_{o}}{T_{o}} = \frac{P_{f}\cdot V_{f}}{T_{f}}[/tex]
Where:
[tex]P_{o}[/tex], [tex]P_{f}[/tex] - Initial and final pressures, measured in torr.
[tex]V_{o}[/tex], [tex]V_{f}[/tex] - Initial and final volumes, measured in liters.
[tex]T_{o}[/tex], [tex]T_{f}[/tex] - Initial and final temperatures, measured in kelvins.
Now, the final volume of the gas is found:
[tex]V_{f} = \left(\frac{T_{f}}{T_{o}}\right)\cdot \left(\frac{P_{o}}{P_{f}} \right)\cdot V_{o}[/tex]
If [tex]T_{o} = 286.15\,K[/tex], [tex]T_{f} = 429.15\,K[/tex], [tex]P_{o} = 568\,torr[/tex], [tex]P_{f} = 897\,torr[/tex] and [tex]V_{o} = 0.35\,L[/tex], the new volume of the gas is:
[tex]V_{f} = \left(\frac{429.15\,K}{286.15\,K} \right)\cdot \left(\frac{568\,torr}{897\,torr} \right)\cdot 0.35\,L[/tex]
[tex]V_{f} = 0.332\,L[/tex]
The new volume of the gas is 0.332 liters.
