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,

\( 4 M_{ X }= M_{ Y }\)

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