The pressure of molecular oxygen in a 10.5 L tank was 1.95 atm at 27.4 degrees C. If the gas is allowed to expand into a larger tank with a volume of 15.5 L at 32.0 degrees C, what would the final pressure of molecular oxygen in the new tank?
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PV = nRT
In the first tank:
n = PV/RT = (1.95 atm) x (10.5 L) / (0.08205746 L atm /K mol) x ( 300.55 K) = 0.830212195 moles
In the second tank:
P = nRT/V = (0.830212195 mol) x (0.08205746 L atm /K mol) x (305.15 K) / (15.5 L) = 1.34 atm
In the first tank:
n = PV/RT = (1.95 atm) x (10.5 L) / (0.08205746 L atm /K mol) x ( 300.55 K) = 0.830212195 moles
In the second tank:
P = nRT/V = (0.830212195 mol) x (0.08205746 L atm /K mol) x (305.15 K) / (15.5 L) = 1.34 atm
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You can use this equation to solve it:
P1 V1 T2 = P2 V2 T1
The 1 is the initial condition and the 2 is the new condition. By the temperature needs to be in Kelvin, so add 273.15 to the Celsius to convert it...
It can be set up like this:
(1.95 atm)(10.5 L)(32 + 273.15 k) = P2 (15.5 L)(27.4 + 273.15 k)
Do the algebra and you get the new pressure to be 1.34 atm.
Hope that helped :)
P1 V1 T2 = P2 V2 T1
The 1 is the initial condition and the 2 is the new condition. By the temperature needs to be in Kelvin, so add 273.15 to the Celsius to convert it...
It can be set up like this:
(1.95 atm)(10.5 L)(32 + 273.15 k) = P2 (15.5 L)(27.4 + 273.15 k)
Do the algebra and you get the new pressure to be 1.34 atm.
Hope that helped :)