what volume will 5.00g of dinitrogen oxide gas occupy at 50.0 degrees celcious and 12.0 atmosphere
Dinitrogen oxide, N2O, is also called nitrous oxide. (There is another nitrogen oxide with two N atoms: N2O4, but that is called nitrogen tetroxide).
The N2O molecular mass is 14+14+16 = 44 g/mole
5g of N2O is n = 5/44 = 0.1136 moles
The ideal gas law
V = nRT/P,
can be used to compute the volume.
Here is another way: At STP, the volume would be 0.1136 x 22.4 = 2.545 liters
At your T and P, multiply 2.545 l by
(1/12)*(323/273) to get the volume in liters. I get 0.25 l
ed to compute the volume.
Here is another way: At STP, the volume would be 0.1136 x 22.4 = 2.545 liters
At your T and P, multiply 2.545 l by
(1/12)*(323/273) to get the volume in liters. I get 0.25 l
To calculate the volume of a gas using the ideal gas law equation, we can use the formula:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/(mol.K))
T = temperature (in Kelvin)
First, let's convert the given temperature of 50.0 degrees Celsius to Kelvin:
T (Kelvin) = T (Celsius) + 273.15
T (Kelvin) = 50.0 + 273.15
T (Kelvin) = 323.15 K
Next, we need to calculate the number of moles of dinitrogen oxide (N2O) gas. To do this, we will use the molar mass of N2O, which is 44.01 g/mol:
n (moles) = mass (grams) / molar mass (g/mol)
n (moles) = 5.00 g / 44.01 g/mol
n (moles) ≈ 0.11352 mol (rounded to 5 decimal places)
Now, we can substitute the known values into the ideal gas law equation:
PV = nRT
V = (nRT) / P
V = (0.11352 mol * 0.0821 L.atm/(mol.K) * 323.15 K) / 12.0 atm
V ≈ 0.767 Liters
Therefore, 5.00 grams of dinitrogen oxide gas will occupy approximately 0.767 Liters at a temperature of 50.0 degrees Celsius and a pressure of 12.0 atmospheres.