Question
Question: The van der waals constant for \({O_2}\) are a= 1.642 atm\[{L^2}mo{l^{ - 2}}\]and b = 0.04 \[Lmo{l^{...
The van der waals constant for O2 are a= 1.642 atmL2mol−2and b = 0.04 Lmol−1. Calculate the temperature at whichO2 gas behaves ideally for a longer range of pressure.
Solution
Look at the second part of the question it says to find out the temperature at which O2 gas (a real gas) behaves ideally for a longer range of pressure, this statement is the very meaning of Boyle's temperature. Hence, we are going to use the concept of Boyle's temperature.
Formula used: TB=bRa
Where, TB= Boyle's temperature
a = correction factor for attractive forces
b = correction factor for the volume of moles
R = gas constant( 0.082 L atm mol− K−)
Complete answer:
Now, simply using the mentioned formula of Boyle's temperature we can determine the required temperature,
So, let's put up the values and calculate,
⇒ TB= 0.04×0.0821.642
We will use the basic arithmetic operations(multiplication followed by division) to solve this,
⇒ TB= 0.003281.642
⇒ TB= 500 K
**Therefore, we can say that at around 500 K O2 gas will behave as an ideal gas.
Additional information:**
1. An ideal gas is a theoretical gas, which follows the ideal gas equation(or law). i.e, PV = nRT , here, P = pressure(force per unit area) of the gas, V = Volume of the gas, n = number of moles, R = Gas constant, T = temperature of the ideal gas.
2. A real gas does not follow the ideal gas equation(or law). However, they obey van der waals gas law, which also gives us the a and b correction factors
3. Van der waals gas equation is (P + V2an2)(V - nb)=nRT , where terms have their usual meaning.
Note:
As the question mentions van der wall and the values associated with this equation, a student might get confused about which formula to use. It is advised to read up the whole question before attempting it.