Question

For a fixed amount of gas, if the absolute temperature of the gas is doubled, what happens to the pressure of the gas?

Which contains more moles of material: 80 grams of Helium gas (He, having atomic weight 4.0 g/mol) or 400 grams of Argon gas (Ar, having atomic weight 40 g/mol)?

Answer 1

Gives $M_{He}=4gm$

$m_{HE}=80gm$

$n_{He}=\frac{m_{He}}{M_{He}}$

$n_{He}=\frac{80}{4}=20mol$

$n_{Ar}=\frac{m_{Ar}}{M_{Ar}}$

$n_{He}=\frac{400}{40}=10mol$

Gas eqution

Pv=nRT

$\frac{P_{He}}{P_{Ar}}=\frac{n_{He}}{n{Ar}}$

$P_{Ar}=\frac{n_{Ar}}{n_{He}}\times P_{He}$

Put value

$P_{Ar}=\frac{10}{20}\times P_{He}$

$P_{Ar}=\frac{1}{2}P_{He}$

Final pressure (Ar -gas) is equal is half of initial pressure (He-gas)