88-L sample of dry air cools from 197 degree celsius to -48 degrees celsius while the pressure is maintained at 2.93 atm. what is the final volume?
(V1/T1) = (V2/T2)
Remember T must be in kelvin.
To find the final volume of the sample of dry air, 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/mol·K)
T = temperature (in Kelvin)
First, let's convert the temperatures from Celsius to Kelvin:
Initial temperature: 197 degrees Celsius = 197 + 273 = 470 K
Final temperature: -48 degrees Celsius = -48 + 273 = 225 K
Now, we can rearrange the ideal gas law equation to solve for the final volume (Vf):
Vf = (Vi x Ti x Pf) / (Pi x Tf)
Where:
Vi = initial volume
Ti = initial temperature (in Kelvin)
Pf = final pressure
Pi = initial pressure
Tf = final temperature (in Kelvin)
Given:
Vi = 88 L (initial volume)
Ti = 470 K (initial temperature)
Pf = Pi = 2.93 atm (constant pressure)
Tf = 225 K (final temperature)
Substituting the given values into the equation, we get:
Vf = (88 L x 470 K x 2.93 atm) / (2.93 atm x 225 K)
Simplifying:
Vf = 88 L x 470 K / 225 K
Vf = 184.89 L
Therefore, the final volume of the sample of dry air is approximately 184.89 liters.