General gas equation | 1st year chemistry chapter 3 important short questions | first year chemistry notes

 General gas equation | 1st year chemistry chapter 3 important short questions | first year chemistry notes

To watch lecture video of this topic on youtube please click here

How the value of the general gas constant ‘R’ can be derived with the help of Avogadro’s law?

Ans. According to Avogadro’s law, the volume of one mole of all the ideal gases at S.T.P are 22.414 dm³. Putting the values of                

P = 1 atm

                                         T = 273.16 K

                                         V = 22.4 dm³

                                          n = 1

we get values of R as

What is the physical meaning of R?

Ans. Physical meanings of value of R (0.0821 dm³ atm K^-1 mol) is that, if we have mole of an ideal gas at 273.16 K and one atmospheric pressure and its temperature is increased by 1 K, then it will absorb 0.0821 dm³ atm of energy.

Calculate the S.I unit of R.

Ans.   P = 101325 Nm^-2

          V = 0.022414 m³

          n = 1 mol

          T = 273.16K

                                        R = 8.3143 JK^-1 mol^-1

How the density of an ideal gas doubles by doubling the pressure or decreasing the temperature on Kelvin scale by ½(half)?

Ans. Density of an ideal gas

Densities of gases are given in g dm^-3 but not in g cm^-3. Why ?

Ans. In gases, particles are very far away from each other as compared  to solids and liquids. So appreciable mass of gas is not present in a smaller volume (1 cm³). Therefore density of gases is expressed in a bigger  unit of volume i.e. g dm^-3 instead of g cm^-3.