Question

What volume will 454 grams of hydrogen occupy at 1.05 atm and 55 degrees Celsius

Answer 1

To calculate the volume of 454 grams of hydrogen gas at 1.05 atm and 55 degrees Celsius, we can use the ideal gas law equation, PV = nRT. First, we convert the temperature to Kelvin and determine the number of moles of hydrogen. Then, we convert the pressure from atm to torr and calculate the volume.

Step-by-step explanation:

To calculate the volume of hydrogen gas, we can use the ideal gas law equation, PV = nRT. Given that the mass of hydrogen is 454 grams and the temperature is 55 degrees Celsius, we need to convert the temperature to Kelvin by adding 273.15. The molar mass of hydrogen is 2 grams/mol. Using these values, we can calculate the number of moles of hydrogen gas:

n = mass / molar mass = 454 g / 2 g/mol = 227 mol

Now, let's convert the pressure from atm to torr:

1 atm = 760 torr

1.05 atm = 1.05 x 760 torr = 798 torr

Finally, we can calculate the volume:

V = (nRT) / P

V = (227 mol x 0.0821 L·atm/(mol·K) x (55 + 273.15) K) / 798 torr

V = 18.74 L

Answer 2

5761.09L

Explanations:

According to the ideal gas equation

`PV=nRT`

where:

• P is the ,pressure ,of the gas in atm

,

• V is the, volume ,in litres

,

• n is the mole of hydrogen gas

,

• R is the ,gas constant

,

• T is the, temperature ,in Kelvin

Given the following parameters

P = 1.05atm

R = 0.082057 L*atm/mol*K

T = 55 + 273 = 328K

Determine the mole of hydrogen

`\begin{gathered} n=\frac{mass}{molar\text{ mass}} \\ n=(454)/(2.02) \\ n=224.75moles \end{gathered}`

Determine the required volume

`\begin{gathered} V=(nRT)/(P) \\ V=(224.75*0.082057*328)/(1.05) \\ V=(6049.15)/(1.05) \\ V=5761.09L \end{gathered}`

Hence the required volume of the gas is approximately 5761.09L