Question

A 500-cc bulb weighs 38.734 grams when evacuated and 39.3135 grams when filled with air at 1 atm pressure and 24$^\circ $C. Assuming that the air behaves as an ideal gas at this pressure, calculate the effective mass of one mole of air.

Answer 1

Hint: Ideal gas is defined as that equation which gives the simultaneous effect of the pressure and temperature on the volume of a gas. Ideal gas equation is written as:
$PV = nRT$
Where pressure is given by P and volume is represented by V and n is the number of moles and R is the gas constant and T is given temperature.

Complete step by step answer:
The volume of the bulb is $500cc$. As we know, $1c{m^3} = 1ml$. Then we can say the volume of the bulb is 500 ml Or 500/1000 ltr. The weight of the bulb when it is evacuated = 38.734 grams and the weight of the bulb when it is filled at 1 atm is 39.3135 grams. The mass occupied by the gas is given as:
39.3135-38.734 = 0.5795 gram
Temperature given is 24$^\circ $C or 273+24=297K. To calculate the molar mass, we will start with ideal gas equation,
\[{\text{PV}} = {\text{nRT}}\] ... (1)
Restructuring the above equation, we get:
\[\dfrac{{{\text{PV}}}}{{{\text{RT}}}} = {\text{n}}\] ... (2)
We know the number of moles is given by the ratio of the given mass to the molecular mass. So we can write $n = \dfrac{w}{M}$
Where w represents the given mass and M is the Molar mass. Replacing the value of n in equation (2), we get:
\[\dfrac{{{\text{PV}}}}{{{\text{RT}}}} = \dfrac{w}{M}\]
\[ \Rightarrow M = \dfrac{{w \times RT}}{{PV}}\]
\[ \Rightarrow M = \dfrac{{0.5795 \times 0.0821 \times 297 \times 1000}}{{1 \times 500}}\]
\[ \Rightarrow M = 28.26g/mol\]

Thus, the effective mass of one mole of a gas is 28.26 g/mol at 1 atmospheric pressure.

Note: The ideal gas equation is very valuable because it gives a very good approximation of gases at high temperature and low pressure. But it is not useful if we need to examine the behavior of gases of undetermined volumes.