- #1
Peter G.
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Hi
1. A car tyre has a volume of 18 x 10-3 m3 and contains air at an excess pressure of 2.5 x 105 N/m2 above atmospheric pressure (1.0 x 105 N/m2) Calculate the volume which the air inside would occupy at atmospheric pressure assuming that its temperature remains unchanged:
My attempt:
PV (Initial) = PV (Final)
(2.5 x 105+1.0 x 105) x 18 x 10-3 / 1.0 x 105[/SUP = V
V = 0.063 m3
2. An oxygen cylinder contains 0.50 kg of gas at a constant pressure of 0.50 MN/m2 and a temperature of 7 degrees Celsius. What mass of oxygen must be pumped into raise the pressure to 3.0 MN/m2 at a temperature of 27 degrees Celsius. If the molar mass of oxgen is 32 x 10-3 kg, calculate the volume of the cylinder
My attempt:
I know this probably can be done in a simpler way but...
First I found the volume with the initial conditions:
V = nRT / P
V = 15.625 x 8.31 x 280 / 500
V = 72.7125
then:
I changed n for mass / molar mass:
m / 32x10-3 = 3000 x 72.7125 / 8.31 x 300
m = 2.8
But since it is how much more oxygen must be pumpted:
m = 2.8-0.5
m = 2.3
Thanks in advance,
Peter G.
(P.S: I never learned moles and I know this question is a bit stupid but I am insecure: The molar mass of carbon dioxide is 44.0 x 10-3 kg. Calculate (a) the number of moles and (b) the number of molecules in 1 kg of the gas:)
I got 22.73 for both.
1. A car tyre has a volume of 18 x 10-3 m3 and contains air at an excess pressure of 2.5 x 105 N/m2 above atmospheric pressure (1.0 x 105 N/m2) Calculate the volume which the air inside would occupy at atmospheric pressure assuming that its temperature remains unchanged:
My attempt:
PV (Initial) = PV (Final)
(2.5 x 105+1.0 x 105) x 18 x 10-3 / 1.0 x 105[/SUP = V
V = 0.063 m3
2. An oxygen cylinder contains 0.50 kg of gas at a constant pressure of 0.50 MN/m2 and a temperature of 7 degrees Celsius. What mass of oxygen must be pumped into raise the pressure to 3.0 MN/m2 at a temperature of 27 degrees Celsius. If the molar mass of oxgen is 32 x 10-3 kg, calculate the volume of the cylinder
My attempt:
I know this probably can be done in a simpler way but...
First I found the volume with the initial conditions:
V = nRT / P
V = 15.625 x 8.31 x 280 / 500
V = 72.7125
then:
I changed n for mass / molar mass:
m / 32x10-3 = 3000 x 72.7125 / 8.31 x 300
m = 2.8
But since it is how much more oxygen must be pumpted:
m = 2.8-0.5
m = 2.3
Thanks in advance,
Peter G.
(P.S: I never learned moles and I know this question is a bit stupid but I am insecure: The molar mass of carbon dioxide is 44.0 x 10-3 kg. Calculate (a) the number of moles and (b) the number of molecules in 1 kg of the gas:)
I got 22.73 for both.