There are two containers filled with gases.In both containers,gases are same temperature and pressure.The first container is 3 L and contains 0.9 moles of gases.
The second container is 1 L.How many moles of gas are there in the second container?
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This can be done by ideal gas law PV = nRT.
n1 = moles in gas 1 = 0.9
n2 = moles in gas 2
`R = 8.314`
`V1 = 3L`
`V2 = 1L`
For both cases pressure and temperature is same.And R is also same for both gases.
`PV = nRT`
`P/(RT) = n/V`
Since P/(RT) is same for both;
`(n1)/(V1) = (n2)/(V2)`
`n2 = (n1V2)/(V1)`
`n2 = 0.9*1/3`
`n2 = 0.3`
So in the other gas there are 0.3 moles.
- Both gasses act as ideal gasses.
To solve for the number of moles of gas in the second container, apply Ideal Gas Law. The formula is:
`PV = nRT`
where P - pressure , V - volume , T - temperature, n - number of moles and R - universal constant (R=8.3145J / mol K ).
The given in the first conatiner are V = 3L and n=9 moles. Substitute these to the formula above.
`P_1(3) = 0.9 (8.3145)T_1`
Then, isolate `P_1` and `T_1` since their values are unknown.
`P_1/T_1 = (0.9(8.3145))/3`
`P_1/T_1 = 2.49435`
And in the second container, the given is V=1 L. Substitute this to the formula of Ideal Gas Law.
Isolate `P_2` and `T_2` too.
`P_2/T_2 = (8.3145n)/1`
`P_2/T_2 = 8.3145n`
Note that the pressure and temperature of the gases inside the two containers are the same. So,
`P_1/T_1 = P_2/T_2`
Then divide both sides by 8.3145 to solve for n.
Hence, there are 0.3 moles of gases in the second container.
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