You collect 6.36 moles of hydrogen gas from your lab experiment involving the decomposition of water by electrolysis. The pressure inside your rigid collection vessel is at 1.00 atm. Another lab group adds their collected gas to yours, giving it an additional 1.28 moles. What is the new pressure inside the container?
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The number of moles of a gas is directly proportional to the pressure of the gas at constant temperature and volume.
This comes from the ideal gas law:
PV=nRT or PV/nT = R, a constant
We can assume constant volume because the container is rigid, and there's no indication that the temperature changed.
When V and T are unchanged they can be left out of the equation so P/n equals a constant. When pressure and number of moles change, P1/n1 = P2/n2.
We can rearrange this to solve for the new pressure: P2 = (P1)(n2)/n1
P1 = 1.00 atm
n1 = 6.36 moles
n2 = 6.36 moles + 1.28 moles = 7.64 moles
P2 = (1.00 atm)(7.64 mol)/(6.36 mol) = 1.20 atm
This answer is consistent with what we expect, a higher pressure because the number of moles of gas increased.
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we can solve this problem with ideal gas equation, which is pv=nRT
where, p=pressure of the container
v=volume of the container
n=number of moles of the gas
R= Universal gas constant
T= temperature in kelvin
If we substitute the values in the first instant in to the equation;(that is before adding the second volume of gas)
1x v =6.36x Rx T----(1)
If we substitute the second set of values( that is after the addition of second gas sample)
p2x v = (6.36+1.28) x Rx T----(2)
note: * since the question doesn't say about a change of temperature
we can assume that the temperature is constant
* Container is said to be rigid so we can assume that the volume is constant.
Divide equation number (1) by (2)
(1x v =6.36x Rx T) ----(1)
____________________
(p2x v = 7.64x Rx T)----(2)
By simplifying these equations you would get
6.36p2 =7.64
p2 = 7.64/6.36
=1.20 atm//
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