A sample of gas exerts a pressure of 8.2atm when continued in a 2.93dm3 container at 20% the number of more of the sample is (R =0.082 atm/dm3 k1 mol1

Answer 1

To solve this problem, we'll need to use the ideal gas law:

PV = nRT

where P is the pressure of the gas, V is the volume of the container, n is the number of moles of the gas, R is the gas constant, and T is the temperature of the gas in Kelvin.

We're given P, V, and R, so we can solve for n:

n = PV/RT

To find T, we need to use the percent composition of the gas. Let's assume that the gas is made up of 20% of some molecule X, and 80% of some other molecule Y. Then the total number of moles in the gas will be:

n_total = n_X + n_Y

We don't know the individual quantities of n_X and n_Y, but we do know that the mole fraction of X is 0.2. This means that:

n_X/n_total = 0.2

We can rearrange this equation to solve for n_X:

n_X = 0.2n_total

Now we can substitute this expression for n_X into our original equation:

n_total = n_X + n_Y
n_total = 0.2n_total + n_Y
0.8n_total = n_Y
n_Y = 0.8n_total

So the mole fraction of Y is 0.8. We can use the mole fractions to calculate the partial pressures of X and Y:

P_X = 0.2P_total
P_Y = 0.8P_total

Now let's substitute these equations into the ideal gas law:

PV_total = n_totalRT

(P_XV + P_YV) = n_totalRT

(0.2P_totalV + 0.8P_totalV) = n_totalRT

P_totalV = n_totalRT

Now we can solve for n_total:

n_total = P_totalV/RT

n_total = (8.2 atm)(2.93 dm^3)/(0.082 atm/dm^3K mol)(293 K)

n_total = 0.979 mol

Since we know that the mole fraction of X is 0.2, we can calculate the number of moles of X:

n_X = 0.2n_total

n_X = 0.196 mol

Therefore, the number of moles of the gas that is made up of molecule X is 0.196 mol.