Question

What volume will 454.5 grams of Hydrogen gas occupy at 1.050 atm and 25.00 ℃? (Don’t forget Hydrogen gas is diatomic - H2.

Answer 1

The volume that will occupy the hydrogen gas is 5288.6486 L

Step-by-step explanation:

Given:

m = mass = 454.5 g

P = pressure = 1.05 atm

T = 25°C = 298 K

Question: What volume will 454.5 grams of Hydrogen gas occupy, V = ?

Before calculating the volume we will need to calculate the number of moles of the gas. The molecular weight of hydrogen is 1 g/mol, but, the hydrogen is diatomic, therefore, will be 2 g/mol

`n=454.5g*(1mol)/(2g) =227.25moles`

Now, we will use the ideal gas equation:

`PV=nRT`

Solving for V:

`V=(nRT)/(P)`

Here

R = constant ideal = 0.082 atm L/mol K

Substituting:

`V=(227.25*0.082*298)/(1.05) =5288.6486L`

Answer 2

The volume that 454.5 grams of gaseous hydrogen will occupy at 1,050 atm and 25.00 ℃ will be 5,288.65 L

Step-by-step explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case it is necessary to know the molar mass to calculate the amount of moles n. The molar mass of H2 is 2 g / mole. Then the following rule of three can be applied: if 2 grams are contained in 1 mole of substance, 454.5 grams in how many moles will they be contained?

`n=(454.5 grams*1 mole)/(2 grams)`

n=227.25 moles

Then, you know:

• P=1.050 atm
• V=?
• n=227.25

• R=0.082
  `(atm*L)/(mol*K)`

• T=25 °C= 298 °K (0°C=273°K)

Replacing:

`1.050 atm*V= 227.25 moles*0.082 (atm*L)/(mol*K) *298K`

Solving:

`V= (227.25 moles*0.082 (atm*L)/(mol*K) *298K)/(1.050 atm)`

V= 5,288. 65 L

The volume that 454.5 grams of gaseous hydrogen will occupy at 1,050 atm and 25.00 ℃ will be 5,288.65 L