What volume would 10.5 g of nitrogen gas, N2 occupy at 200.K and 2.02 atm.

convert 10.5g N2 to moles.

moles= 10.5/28 as I remember, check it.

V= nRT/P check your R units, to make certain it is in atm

To find the volume of nitrogen gas, N2, we can use the ideal gas law equation:

PV = nRT

Where:
P is the pressure of the gas (in atm)
V is the volume of the gas (in liters)
n is the number of moles of gas
R is the ideal gas constant (0.0821 L.atm/mol.K)
T is the temperature of the gas (in Kelvin)

First, we need to convert the given temperature from Celsius to Kelvin:
200 K = 200 + 273.15 = 473.15 K

We also need to convert the given mass of nitrogen gas into moles. To do this, we need the molar mass of nitrogen gas, which is 28.02 g/mol. Divide the given mass (10.5 g) by the molar mass to get the number of moles:

n = 10.5 g / 28.02 g/mol = 0.374 mol

Now, we can plug the values into the ideal gas law equation:

PV = nRT

(2.02 atm) * V = (0.374 mol) * (0.0821 L.atm/mol.K) * (473.15 K)

Simplifying the equation, we get:

2.02 * V = 0.374 * 0.0821 * 473.15

2.02V = 12.03

Finally, we can solve for V:

V = 12.03 / 2.02

V ≈ 5.96 L

Therefore, 10.5 g of nitrogen gas (N2) would occupy approximately 5.96 liters at 200 K and 2.02 atm.

To solve this problem, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(K·mol))
T = temperature (in Kelvin)

We have the pressure (P = 2.02 atm), temperature (T = 200 K) and the molar mass of N2 gas (28 g/mol). To find the volume (V), we first need to determine the number of moles (n) using the mass of the gas.

First, find the number of moles (n):

n = mass / molar mass

Given that mass = 10.5 g and the molar mass of N2 = 28 g/mol:

n = 10.5 g / 28 g/mol

n ≈ 0.375 mol

Now we can substitute the known values into the ideal gas law equation and solve for V:

PV = nRT

(2.02 atm) V = (0.375 mol) (0.0821 L·atm/(K·mol)) (200 K)

V = (0.375 mol) (0.0821 L·atm/(K·mol)) (200 K) / (2.02 atm)

V ≈ 7.43 L

Therefore, 10.5 g of nitrogen gas N2 would occupy approximately 7.43 liters at a temperature of 200 K and a pressure of 2.02 atm.