This can be done by ideal gas law PV = nRT.

Let;

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...

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This can be done by ideal gas law PV = nRT.

Let;

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.*

*Assumption*

*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.

`P_2(1)=n(8.3145)T_2`

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`

`2.49435=8.3145n`

Then divide both sides by 8.3145 to solve for n.

`2.49435/8.3145=(8.3145n)/8.3145`

`n =0.3`

**Hence, there are 0.3 moles of gases in the second container.**