how many liters are occupied by 0.908 mol of nitrogen at 108 degrees C and 0.858 atm pressure?
use the ideal gas law by solving for volume v=nRT/p= (.908mol)(.08206)(381K)/(.858atm)=???
R=.08206 latm/molK
Temperature needs to be converted to Kelvin K=108+273
To find the volume occupied by a gas given its temperature, pressure, and number of moles, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/(K·mol))
T = Temperature (in Kelvin)
First, we need to convert the temperature from Celsius to Kelvin. The formula for converting Celsius to Kelvin is:
T (Kelvin) = T (Celsius) + 273.15
So, let's convert the temperature of 108 degrees Celsius to Kelvin:
T (Kelvin) = 108 + 273.15 = 381.15 K
Now we can rearrange the ideal gas law equation to solve for the volume (V):
V = (nRT) / P
Let's plug in the known values:
n = 0.908 mol (number of moles)
R = 0.0821 L·atm/(K·mol) (gas constant)
T = 381.15 K (temperature)
P = 0.858 atm (pressure)
V = (0.908 mol * 0.0821 L·atm/(K·mol) * 381.15 K) / 0.858 atm
After performing the calculation, we find that the volume occupied by 0.908 mol of nitrogen at 108 degrees Celsius and 0.858 atm pressure is approximately 34.06 liters.