Question

The volume of a gas at $ {0^ \circ }C $ and 760 mm pressure is $ 22.4cc $. The number of molecules present in this volume is
1) $ {10^{ - 3}}{N_A} $
2) $ {10^{ - 4}}{N_A} $
3) $ {10^{ - 5}}{N_A} $
4) $ {10^{ - 2}}{N_A} $

Answer 1

Hint: At first calculate the number of moles by dividing the product of pressure in atmosphere and volume in litre by the product of ideal or universal gas constant and temperature in Kelvin. Then, the number of molecules is calculated by multiplying the number of moles and Avogadro Number.

Complete Step by step answer:
The ideal gas law is the equation of state of a hypothetical ideal gas. It is also known as the general gas equation. It is a good approximation of the behaviour of many gases under many conditions. The ideal gas equation states that the product of pressure $ \left( P \right) $ and volume $ \left( V \right) $ is equal to the product of amount of substance $ \left( n \right) $, ideal gas constant $ \left( R \right) $ and temperature $ \left( T \right) $ i.e., $ PV = nRT $.
The volume $ \left( V \right) $ of a gas is 22.4 cc at temperature $ \left( T \right) $ is $ {0^ \circ }C $ and pressure $ \left( P \right) $ is 760 mm. 1 atmospheric pressure is equal to 760 mm of height of mercury level i.e., $ 1atm = 760mm $. Now, we’ll convert temperature from $ ^ \circ C $ to Kelvin and volume from cc (cubic centimetre) to L (litre).
 $ 1K = {273^ \circ }C $ and $ 1cc = {10^{ - 3}}L $.
Hence, $ {0^ \circ }C = 273 + 0 $ i.e., 273 K and $ 22.4cc = 22.4 \times {10^{ - 3}}L $

Now, we’ll put all the values in the formula $ PV = nRT $.
 $ 1 \times 22.4 \times {10^{ - 3}} = n \times 0.0821 \times 273 $ where $ R = 0.0821 $ $ Latm $ $ mo{l^{ - 1}}{K^{ - 1}} $
 $ \Rightarrow 22.4 \times {10^{ - 3}} = 22.4n $
 $ \Rightarrow n = \dfrac{{22.4 \times {{10}^{ - 3}}}}{{22.4}} $
 $ \Rightarrow n = {10^{ - 3}} $ i.e., number of moles $ \left( n \right) = {10^{ - 3}} $
Thus, the number of molecules $ \left( N \right) = n \times {N_A} $ where $ {N_A} $ is Avogadro Number.
 $ N = {10^{ - 3}}{N_A} $.

Therefore, option 1 is correct.

Note: Remember the equation of ideal gas law and also do not forget to convert temperature from degree Celsius to Kelvin, pressure in mm of Hg (mercury) to atmospheric pressure and volume in cubic centimetre to litre.