Question

Suppose two 200.0-L tanks are to be filled separately with the gases helium and hydrogen. What mass of each gas is needed to produce a pressure of 135 atm in its respective tank at 24C?

Answer 1

Answer:

Mass of helium  = 4426.9524 g

Mass of hydrogen gas = 2213.4762 g

Explanation:

Pressure = 135 atm

Temperature = 24 °C

Volume = 200 L

Number of moles = ?

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (24 + 273.15) K = 297.15 K  

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

135 atm × 200 L = n × 0.0821 L.atm/K.mol × 297.15 K  

⇒n = 1106.7381 moles

For helium gas:

Molar mass = 4 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$1106.7381= \frac{Mass}{4}$

$Mass= 4426.9524\ g$

For hydrogen gas:

Molar mass = 2 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$1106.7381= \frac{Mass}{2}$

$Mass= 2213.4762\ g$