Question

0.820 mole of hydrogen gas has a volume of 2.00 l at a certain temperature and pressure. what is the volume of 0.125 mol of this gas at the same temperature and pressure?

Answer 1

Volume is 0.305 L

Step-by-step explanation:

The ideal gas law is:

PV = nRT

We have 0.820 mole of
`H_(2)` gas at temperature and pressure of T1 and P1 and a volume of 2.00L.

P1(2.0L)=(0.820 mole)*R*(T1)

A second sample has 0.125 moles of
`H_(2)` at the same temperature and pressure and unknown volume:

P1(V2) = (0.125 mole)*R*(T1)

We want to find the volume, V2:

V2 = ((0.125 mole)*R*(T1))/P1

From the first sample, rearrange to find P1:

P1(2.0L) = (0.820 mole)*R*(T1)

P1 = ((0.820 mole)*R*(T1))/(2.0L)

Now use this value of P1 in the equation for the second sample:

V2 = ((0.125 mole)*R*(T1))/P1

V2 = ((0.125 mole)*R*(T1))/(((0.820 mole)*R*(T1))/(2.0L))

V2 = ((0.125 mole)*R*(T1))/(((0.820 mole)*R*(T1))/(2.0L)) The R and T1 terms cancel:

V2 = ((0.125 mole))/((0.820 mole)/(2.0L))

V2 = ((0.125)/(0.820)) * (2.0L)

V2 = (0.1524)*(2.0 L)

V2 = 0.305 L