Answer:
[tex]V_{2} = 8.92 L[/tex]
Explanation:
We have the equation for ideal gas expressed as:
PV=nRT
Being:
P = Pressure
V = Volume
n = molar number
R = Universal gas constant
T = Temperature
From the statement of the problem I infer that we are looking to change the volume and the pressure, maintaining the temperature, so I can calculate the right side of the equation with the data of the initial condition of the gas:
[tex]P_{1} V_{1} =nRT[/tex]
[tex]320Kpa*0.003m^{3} =nRT[/tex]
[tex]1000L = 1m^{3}[/tex]
So
[tex]nRT= 0.96[/tex]
Now, as for the final condition:
[tex]P_{2}V_{2}=nRT[/tex]
[tex]P_{2} V_{2} =0.96[/tex]
clearing[tex]V_{2}[/tex]
[tex]V_{2} =\frac{0.96}{P_{2} }[/tex]
[tex]V_{2} =0.00872m_{3}[/tex]
[tex]V_{2} = 8.92 L[/tex]