Answer:
The volume occupied by fluorine gas, when the pressure is doubled and temperature is increased to 400 K is 600 ml.
Explanation:
As per ideal gas law, the pressure of any gas is inversely proportional to the volume occupied by the gas. Similarly, the volume of the gas is directly proportional to the temperature of the gas molecules. So the ideal gas equation is [tex]PV = nRT[/tex]. Here P is pressure, V is volume, n is the no.of moles, R is gas constant and T is temperature.
So in this case, the fluorine gas occupies 810 mL of volume ([tex]V_{1}[/tex]) at 270 K temperature ([tex]T_{1}[/tex]) and pressure of about 1 atm ([tex]P_{1}[/tex]). If the pressure is doubled, then the new pressure will be [tex]P_{2}=2*P_{1}=2 atm[/tex] and temperature is increased to 400 K. As the no.of moles will be constant, then
[tex]\frac{P_{1} V_{1} }{P_{2} V_{2} }=\frac{T_{1} }{T_{2} }[/tex]
Thus,[tex]\frac{V_{1} }{V_{2} }=\frac{P_{2}T_{1} }{P_{1} T_{2} }[/tex]
So,
[tex]\frac{810}{V_{2} }=\frac{2*270}{1*400}[/tex]
[tex]V_{2}=\frac{(810*20)}{27}=600 mL[/tex]
Thus, the new volume occupied by fluorine gas on increase in pressure and temperature is 600 mL.
