## Dalton’s law of partial pressure is not applicable to, at normal conditions

Problem : Dalton’s law of partial pressure is not applicable to, at normal conditions

(A) H2 and N2 mixture

(B) H2 and Cl2 mixture

(C) H2 and CO2 mixture

(D) H2 and O2 mixture

Ans: (B)

Solution: H2 and Cl2 reacts to form HCl; Dalton’s law of partial pressure is valid only for the gases which don’t react at ordinary conditions

## Helium atom is two times heavier than a hydrogen molecule. At 298 K, the average kinetic energy…

Problem: Helium atom is two times heavier than a hydrogen molecule. At 298 K, the average kinetic energy of a Helium atom is

(A) two times that of hydrogen molecule

(B) same as that of a hydrogen molecule

(C) four times that of a hydrogen molecule

(D) half that of a hydrogen molecule

Ans: (B)

Solution: The average kinetic energy of an atom is given as $\large \frac{3}{2}k T$

Hence , It does not depend on mass of the atom.

## If for two gases of molecular weights MA and MB at temperature TA and TB …

Problem : If for two gases of molecular weights MA and MB at temperature TA and TB, TA MB = TB MA , then which property has the same magnitude for both the gases.

(A) density

(B) pressure

(C) KE per mol

(D) Vrms

Ans: (D)

Sol:(i) Density of gas $\large \rho = \frac{P M}{R T}$

Since $\large \frac{M_B}{T_B} = \frac{M_A}{T_A}$

Hence , at the same pressure $\large \rho_A = \rho_B$

But if pressure is different then $\large \rho_A \ne \rho_B$

(ii) Pressure of the gases would be equal if their densities are equal other wise not.

(iii) KE per mol $\large = \frac{3}{2}R T$

It will be different for the two gases.

(iv) $\large V_{rms} = \sqrt{\frac{3RT}{M}}$

Since $\large \frac{T_B}{M_B} = \frac{T_A}{M_A}$

Vrms of A = Vrms of B

## At 100 degree C and 1 atm, if the density of liquid water is 1.0 g/cc and that of water vapour is 0.0006 g/cc…

Problem : At 100 °C and 1 atm , if the density of liquid water is 1.0 g/cc and that of water vapour is 0.0006 g/cc, then the volume occupied by water molecule in one litre of steam at that temperature is

(A) 6 cc

(B) 60 cc

(C) 0.6 cc

(D) 0.06 cc

Ans: (C)

Solution: Mass of 1 lt water vapour = V × d = 1000 × 0.0006 = 0.6 g

Volume of liquid water = 0.6/1 = 0.6 cc

## The circulation of blood in human body supplies O2 and releases CO2. the concentration…

Problem : The circulation of blood in human body supplies O2 and releases CO2. the concentration of O2 and CO2 is variable but on an average, 100 ml blood contains 0.02 g of O2 and 0.08 g of CO2. The volume of O2 and CO2 at 1 atm and at body temperature 37 °C, assuming 10 lt blood in human body, is

(A) 2 lt, 4 lt

(B) 1.5 lt, 4.5 lt

(C) 1.59 lt, 4.62 lt

(D) 3.82 lt, 4.62 lt

Ans: (C)

Solution: 100 ml blood has 0.02 g O2 and 0.08 g CO2

10,000 ml blood has 2 g O2 and 8 g CO2

using PV = nRT

for O2, $\large 1 \times V_{O_2} = \frac{2}{32} \times 0.0821 \times 310$

= 1.59 litre

for CO2 , $\large 1 \times V_{CO_2} = \frac{8}{44} \times 0.0821 \times 310$

= 4.62 litre