what is the pressure exerted by a 0.500 mol sample of nitrogen in a 10.0 L container at 20 degrees C?
To find the pressure exerted by a gas sample, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin.
Let's break down the given information:
- Number of moles (n) = 0.500 mol
- Volume (V) = 10.0 L
- Temperature (T) = 20 degrees C
(Note that the ideal gas law equation requires temperature to be in Kelvin, so we need to convert Celsius to Kelvin.)
To convert Celsius to Kelvin, we use the formula: Kelvin = Celsius + 273.15.
So, the temperature in Kelvin would be: T = 20 + 273.15 = 293.15 K.
Now, we can plug these values into the ideal gas law equation:
PV = nRT
P * 10.0 = 0.500 * 0.0821 * 293.15
Simplifying the equation:
P * 10.0 = 12.091
Dividing both sides of the equation by 10.0:
P = 1.2091
Therefore, the pressure exerted by the 0.500 mol sample of nitrogen in the 10.0 L container at 20 degrees C is approximately 1.2091 atm.
To find the pressure exerted by a 0.500 mol sample of nitrogen in a 10.0 L container at 20 degrees C, 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)
Step 1: Convert the temperature from Celsius to Kelvin.
To convert Celsius to Kelvin, add 273.15 to the temperature in Celsius.
T = 20 degrees C + 273.15 = 293.15 K
Step 2: Plug the values into the Ideal Gas Law equation.
PV = nRT
P * 10.0 L = 0.500 mol * 0.0821 L·atm/K·mol * 293.15 K
Step 3: Solve for P.
P = (0.500 mol * 0.0821 L·atm/K·mol * 293.15 K) / 10.0 L
P = 12.087 atm
Therefore, the pressure exerted by a 0.500 mol sample of nitrogen in a 10.0 L container at 20 degrees C is approximately 12.087 atm.