Question

solve the question in the attachment !!​

Answer 1

GiveN :

$\sf{For\ Gas\ in\ 1st\ container} \begin{cases} \sf{Mass\ =\ m_1} \\ \\ \sf{Pressure\ =\ P_1} \end{cases} \\ \\ \\ \\ \sf{For\ gas\ in\ 2nd\ Container} \begin{cases} \sf{Mass\ =\ m_2} \\ \\ \sf{Pressure\ =\ P_2} \end{cases}$

To FinD :

• Common Pressure , if temperature remains constant

SolutioN :

We've Ideal gas equation :

$\implies \boxed{\boxed{\sf{PV\ =\ nRT}}} \\ \\ \\ \implies \sf{PV\ =\ \dfrac{m}{M} RT}$

Here,

• n = Number of moles
• P is pressure
• T is temperature
• R is gas constant
• M is molar mass of gas
• m is given mass

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$\underbrace{\sf{For\ Gas\ in\ first\ container}} \\ \\ \\ \implies \sf{P_1 V_1\ =\ \dfrac{m_1}{M} RT} \\ \\ \\ \implies \sf{V_1\ =\ \dfrac{m_1 RT}{MP_1} \: \: \: \: \: \: \: \: \: \: ...(1)}$

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$\underbrace{\sf{For\ Gas\ in\ second\ container}} \\ \\ \implies \sf{P_2 V_2\ =\ \dfrac{m_2}{M} RT} \\ \\ \\ \implies \sf{V_2\ =\ \dfrac{m_2 RT}{MP_2} \: \: \: \: \: \: \: \: \: ...(2)}$

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Now, If gasses are putted in contact with each other then :

Add (1) and (2)

$\implies \sf{V_1\ +\ V_2\ =\ \dfrac{m_1RT}{MP_1}\ +\ \dfrac{m_2RT}{MP_2}} \\ \\ \\ \implies \sf{V_1\ +\ V_2\ =\ \dfrac{(m_1\ +\ m_2)RT}{M}\ \times\ \dfrac{1}{P_1\ +\ P_2}} \\ \\ \\ \implies \sf{V_1\ +\ V_2\ =\ \dfrac{(m_1\ +\ m_2) RT}{M}\ \times\ \dfrac{1}{P_{eff}}}$

Now, put value of $\sf{V_1\ and\ V_2\ from\ (1)\ and\ (2)}$

$\implies \sf{\bigg( \dfrac{m_1}{MP_1} RT \bigg)\ +\ \bigg( \dfrac{m_2}{MP_2} RT \bigg)\ =\ \dfrac{(m_1\ +\ m_2) RT}{MP_{eff}}} \\ \\ \\ \implies \sf{\cancel{\dfrac{RT}{M}} \bigg( \dfrac{m_1}{P_1}\ +\ \dfrac{m_2}{P_2} \bigg) \ =\ \dfrac{(m_1\ +\ m_2) \cancel{RT}}{\cancel{M} P_{eff}}} \\ \\ \\ \implies \sf{\dfrac{m_1}{P_1} \ +\ \dfrac{m_2}{P_2}\ =\ \dfrac{m_1\ +\ m_2}{P_{eff}}} \\ \\ \\ \implies {\underline{\boxed{\sf{P_{eff}\ =\ \dfrac{(m_1\ +\ m_2)P_1 P_2}{P_2m_1\ +\ P_1m_2} }}}}$