Ncert Exercises & Solutions

5.5 Pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.

For ideal gas A, the ideal gas equation is given by,

p_{A} V = n_{A} R T . . . . . (i)

Where,

p_{A} a n d n_{A}

represent the pressure and number of moles of gas A.

For ideal gas B, the ideal gas equation is given by,

p_{B} = n_{B} R T . . . . . (i i)

Where,

p_{B} a n d n_{B}

represent the pressure and number of moles of gas B.

(V and T are constants for gases A and B)

From equation (i), we have

p_{A} V = \frac{m_{A}}{M_{A}} R T \Rightarrow \frac{p_{A} M_{A}}{m_{A}} = \frac{R T}{V} . . . . . . (i i i)

From equation (ii), we have

p_{B} V = \frac{m_{B}}{M_{B}} R T \Rightarrow \frac{p_{B} M_{B}}{m_{B}} = \frac{R T}{V} . . . . . . (i i i)

Where, MA and MB are the molecular masses of gases A and B respectively.

\frac{p_{A} M_{A}}{m_{A}} = \frac{p_{B} M_{B}}{m_{a}} . . . . . . . . . (v)

Given,

m_{A} = 1 g

p_{A} = 2 b a r

m_{B} = 2 g

p_{B} = (3 - 2) = 1 b a r

(Since total pressure is 3 bar)

Substituting these values in equation (v), we have

\frac{2 \times M_{A}}{1} = \frac{1 \times M_{B}}{2}

\Rightarrow 4 M_{A} = M_{B}

Thus, a relationship between the molecular masses of A and B is given by

4 M_{A} = M_{B}

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