# 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...

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?

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.**

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.*