A gas at 250K and 15atm has a molar volume smaller than calculated from perfect gas law
Calculate:
a) the compression factor under these conditions
b) the molar volume of gas
<Note>
I find the problem from the testbook: Physical Chemistry 10th by Atkins.
In this book, It gives "gas at 250 K and 15 atm has a molar volume 12 per cent smaller than that calculated from perfect gas law", so I used the condition"12%".
If your professer deleted the condition, please use x stand for 12%. Of course the solution is going to exist unknown number "x".
1st think that: perfect gas law means PV = nRT, and molar volume is V/n = Vm (m is subscript actuall)
so, it can be PVm = RT
Z = PVm/RT = 1 for perfect gas (or ideal gas)
now, we know that" molar volume smaller than calculated from perfect gas law "
so Vm(gas) = (100-12)%Vm=0.88Vm
a) Z= PVm(gas)/RT
= P(0.88Vm)/RT = 0.88 ~ solved.
[Hint: Z is dimensionless, so is not unit]
b) you can look for some data table to know R= 0.08206 (atm)(L)/(K)(mol);
or you can use the SI unit:
R= 8.314 J/(K)(mol) , and change unit by yourself.
from a)
Z= 0.88= PVm(gas)/RT
= (15)(Vm(gas))/(0.082)(250)
=0.73 (Vm(gas))
calculated finally,
Vm(gas)= 0.88/0.73
= 1.2 (L/mol) ~ solved.
Thank you
a) Well, the gas must be feeling pretty "compressed" under these conditions! To calculate the compression factor, we can use the formula Z = PV/RT, where Z is the compression factor, P is the pressure, V is the volume, R is the ideal gas constant, and T is the temperature. Since we're given the pressure, temperature, and assuming it's an ideal gas, we can find the compression factor.
From the ideal gas law, PV = nRT, we can solve for V to get V = (nRT)/P.
Since the volume is smaller than predicted, we know that the compression factor, Z, must be less than 1. In other words, it's feeling quite compressed!
b) To calculate the molar volume of the gas, we can use the formula V = (nRT)/P, where V is the molar volume, n is the number of moles, R is the ideal gas constant, T is the temperature, and P is the pressure.
Since we're not given the number of moles, we can't calculate the exact molar volume. But we can use the ideal gas law as an approximate guide.
To calculate the compression factor and molar volume of a gas under the given conditions, we need to use the ideal gas law equation and the van der Waals equation for real gases.
a) The compression factor (Z) can be calculated using the van der Waals equation:
Z = (Vreal gas) / (Videal gas)
The ideal gas law equation is given as:
PV = nRT
where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
Since we're given the temperature (T = 250K) and pressure (P = 15 atm), we can rearrange the ideal gas law equation to solve for the molar volume (V) of the ideal gas:
Videal gas = (nRT) / P
To calculate the molar volume (Vreal gas) of the real gas, we'll use the van der Waals equation:
(P + a(n/V^2))(V - nb) = nRT
where:
a and b are constants specific to the gas.
Rearranging the van der Waals equation to solve for Vreal gas:
Vreal gas = (nRT + nb) / (P + a(n/V^2))
Now we can substitute the given values into the equations to find the values of Z (compression factor) and Vreal gas (molar volume of the gas).
b) Since we're given a temperature (T = 250K) and pressure (P = 15 atm), we can substitute these values into the ideal gas law equation to calculate the molar volume (Videal gas):
Videal gas = (nRT) / P
Note: For part b, you didn't specify the number of moles (n) of gas. So, if you provide the value of moles, we can calculate the molar volume.