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Question: How many air molecules are present at room temperature \( ({27^\circ }C) \) , \( 1atm \) pressure?...

How many air molecules are present at room temperature (27C)({27^\circ }C) , 1atm1atm pressure?

Explanation

Solution

In order to this question, to know the air molecules that are present at room temperature of given pressure, we will first convert the given temperature into Kelvin and then apply Ideal Gas Law.

Complete step by step solution:
Given temperature =27C=300K= {27^\circ }C = 300K
And Pressure =1atm= 1atm
Now, from the ideal gas law- PV=nRTPV = nRT , we have:
nV=PRT=1atm0.082L×atmmol×K×300K=0.046molL\because \dfrac{n}{V} = \dfrac{P}{{RT}} = \dfrac{{1atm}}{{0.082\dfrac{{L \times atm}}{{mol \times K}} \times 300K}} = 0.046\dfrac{{mol}}{L}
Since 1mol=6.02×1023molecules1mol = 6.02 \times {10^{23}}molecules
So, we will have 2.45×1022molecules/L2.45 \times {10^{22}}molecules/L .

Note:
An ideal gas contains molecules of a negligible size that have an average molar kinetic energy that depends only on temperature. Intermolecular forces and molecular size are not considered by the Ideal Gas Law. The Ideal Gas Law applies best to monatomic gases at low pressure and high temperature.