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

Asked by Pragya Singh | 1 year ago |  79

##### Solution :-

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

$$p_{X}V = n_{X}RT$$……(1)

Where $$p_{X}$$​ and $$n_{X}$$​ represents the pressure and number of moles of gas X.

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

$$p_{Y}V = n_{Y}RT$$……(2)

Where, $$p_{Y}$$​ and $$n_{Y}$$​ represent the pressure and number of moles of gas Y.

[V and T are constants for gases X and Y]

From equation (1),

$$p_{ X }V = \dfrac{m_{ X }}{M_{ X }}$$

$$\dfrac{p_{ X }M_{ X }}{m_{ X }}= \dfrac{ R T}{ V }$$​ ……(3)

From equation (2),

$$p_{ Y }V =\dfrac{m_{ Y }}{M_{ Y }}=RT$$

$$\dfrac{p_{ Y }M_{ Y }}{m_{ Y }}= \dfrac{ R T}{V}$$ …… (4)

Where, $$M_{ X }$$ and $$M_{ Y }$$​ are the molecular masses of gases X and Y respectively.

Now, from equation (3) and (4),

$$\dfrac{p_{ X }M_{ X }}{m_{ X }}= \dfrac{p_{ Y }M_{ Y }}{m_{ Y }}$$​​ ….. (5)

Given,

$$m_{ X }$$= 1 g

$$p_{ X }$$= 2 bar

$$m_{ Y }$$= 2 g

$$p_{ Y }$$ = (3 – 2) = 1 bar (Since total pressure is 3 bar)

Substituting these values in equation (5),

$$\dfrac{2 \; \times \; M_{X} }{1}= \dfrac{1 \; \times \; M_{Y} }{2}​​$$

$$4 M_{ X } = M_{ Y }$$

Therefore, the relationship between the molecular masses of X and Y is,

$$4 M_{ X }= M_{ Y }$$

Answered by Abhisek | 1 year ago

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