If 20.0 g of N2 has a volume of 4.00 L and a pressure of 6.0 atm, what is its temperature?
Use PV = nRT
To determine the temperature of a gas, we can use the Ideal Gas Law equation, which states:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's rearrange the equation to solve for temperature:
T = PV / (nR)
Now, we need to gather the necessary values to plug into the equation. Given:
P = 6.0 atm
V = 4.00 L
n = ?
R = 0.0821 L·atm/(mol·K) (the ideal gas constant)
We have the pressure (P) and volume (V), but we need to determine the number of moles of N2 (n) to proceed. To do this, we can use the molar mass of N2, which is:
Molar Mass (N2) = 2 x Atomic Mass (N) = 2 x 14.01 g/mol = 28.02 g/mol
Given that we have 20.0 g of N2, we can calculate the number of moles:
n = mass / molar mass = 20.0 g / 28.02 g/mol ≈ 0.714 mol
Now that we have all the values, we can substitute them into the equation:
T = (6.0 atm) x (4.00 L) / (0.714 mol) x (0.0821 L·atm/(mol·K))
After calculation, we find:
T ≈ 247.5 K
Therefore, the temperature of 20.0 g of N2 with a volume of 4.00 L and a pressure of 6.0 atm is approximately 247.5 Kelvin.