Question

If the oxygen is replaced by helium at the same temperature, how many kilograms of the latter will be needed to produce a gauge pressure of 5.40 atm?

Answer 1

This is an incomplete question, here is a complete question.

A tank contains 30.0 kg of O₂ gas at gauge pressure of 8.20 atm. If the oxygen is replaced by helium at the same temperature. How many kilograms of the latter will be needed to produce a gauge pressure of 7.00 atm?

Answer : The mass of latter needed will be, 2.47 kg

Explanation :

Using ideal gas equation:

$PV=nRT\\\\PV=\frac{w}{M}RT\\\\PVM=wRT$

where,

P = pressure of gas

V = volume of gas

T = temperature of gas

R = gas constant = 0.0821 L.atm/mole.K

w = mass of gas

M = molar mass of gas

The expression for two gases at same temperature, volume of gas will be:

$\frac{P_{O_2}M_{O_2}}{P_{He}M_{He}}=\frac{w_{O_2}}{w_{He}}$

where,

$P_{O_2}$ = pressure of O₂ gas = 8.20 atm

$P_{He}$ = pressure of He gas = 5.40 atm

$M_{O_2}$ = molar mass of O₂ gas = 32 g/mol

$M_{He}$ = molar mass of He gas = 4 g/mol

$w_{O_2}$ = mass of O₂ gas = 30.0 kg

$w_{He}$ = mass of He gas = ?

Now put all the given values in the ideal gas equation, we get:

$\frac{(8.20atm)\times 32g/mol)}{(5.40atm)\times 4g/mol)}=\frac{30.0kg}{w_{He}}$

$w_{He}=2.47kg$

Therefore, the mass of latter needed will be, 2.47 kg