Question

The specific heat of air at constant pressure is $ 1.005kJk{{g}^{-1}}{{k}^{-1}} $ and the specific heat of air at constant volume is $ 0.718kJk{{g}^{-1}}{{k}^{-1}} $ if the universal constant is $ 8.314kJkmo{{l}^{-1}}{{k}^{1}} $ . Find the molecular weight of air?
(A) $ \text{28}\text{.97} $
(B) $ \text{24}\text{.6} $
(C) $ \text{22}\text{.8} $
(D) $ \text{19}\text{.6} $

Answer 1

Hint :We know that the mass of a molecule of a substance is measured in molecular weight which is dependent on $ 12 $ as atomic weight of $ carbon-12 $ in practice. It’s determined by adding the atomic weights of the atoms that make up a substance molecular formula.

Complete Step By Step Answer:
As we know, the symbol $ R $ stands for the molar gas constant (also known as the gas constant, universal gas constant or ideal gas constant.) it’s the major equivalent of the Boltzmann constant. But in terms of energy per temperature increment per mole, i.e. the pressure-volume product, rather than energy per temperature increment per atom.
The values given here are the specific heat of air at constant volume $ {{C}_{v}}=0.718kJk{{g}^{-1}}{{k}^{-1}} $
Also the specific heat of air at constant pressure: $ {{C}_{P}}=0.718kJk{{g}^{-1}}{{k}^{-1}} $
Universal gas constant $ R=8.314kJkmo{{l}^{-1}}{{k}^{1}} $
Also, we can say that
 $ R=\dfrac{8.314}{M}KJK{{g}^{-1}}{{K}^{-1}}(\because n=\dfrac{x}{M}) $
Here, we know that, $ {{C}_{P}}-{{C}_{V}}=R $
 $ 1.005-0.718=\dfrac{8.314}{M} $
 $ M=\dfrac{8.314}{0.287}=28.97 $
Hence, the correct answer is option A i.e. $ 28.97 $ .

Note :
Remember that the amount of heat per unit mass needed to increase the temperature by one degree Celsius is known as specific heat. The relationship between heat and temperature change is commonly expressed as follows, where c is the real heat. The equation for finding the molecular weight when specific heat at constant pressure, specific heat at volume and $ R $ is as follows:
 $ M=\dfrac{R}{{{C}_{p}}-{{C}_{V}}} $