Chemistry please help.
A 20.0-L gas cylinder contains 4.80 g H2 at 25 degree celcius. what is the pressure of this gas?
Use PV = nRT. Don't forget T must be in kelvin. Note the correct spelling of celsius.
To find the pressure of the gas, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure of the gas (in atm)
V = volume of the gas (in liters)
n = number of moles of the gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature of the gas (in Kelvin)
First, we need to convert the temperature from Celsius to Kelvin. To do this, we add 273.15 to the Celsius value:
25°C + 273.15 = 298.15 K
Next, we need to determine the number of moles of H2 in the cylinder. We can use the molar mass of H2 to convert the given mass to moles:
Molar mass of H2 = 2.02 g/mol
Number of moles (n) = mass (in grams) / molar mass (in grams/mol)
= 4.80 g / 2.02 g/mol
Now we can calculate the pressure using the ideal gas law:
P = nRT / V
Substituting the given values into the equation:
P = (number of moles) x (ideal gas constant) x (temperature) / volume
P = (4.80 g / 2.02 g/mol) x (0.0821 L·atm/mol·K) x (298.15 K) / 20.0 L
Now we can calculate the pressure:
P = (2.376 mol) x (0.0821 L·atm/mol·K) x (298.15 K) / 20.0 L
To find the pressure of the gas in the cylinder, we can make use of the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 25 + 273.15 = 298.15 K
Next, we need to calculate the number of moles of H2 gas using the given mass and the molar mass of H2:
molar mass of H2 = 2 g/mol
n = mass / molar mass
n = 4.80 g / 2 g/mol
n = 2.40 mol
Now we can substitute the values into the ideal gas law equation and solve for the pressure:
PV = nRT
P(20.0 L) = (2.40 mol)(0.0821 L·atm/mol·K)(298.15 K)
P = (2.40 mol)(0.0821 L·atm/mol·K)(298.15 K) / 20.0 L
P ≈ 2.48 atm
Therefore, the pressure of the gas in the cylinder is approximately 2.48 atm.