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NCERT Class XI Chemistry States of Matter Solutions

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Question : 5 of 23
Marks: +1, -0
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.
Solution:  
For gas A, PAVP_A V = mAMA\frac{m_A}{M_A} RT ... (1)
For gas B, PBVP_B V = mBMB\frac{m_B}{M_B} RT ... (ii)
Dividing equation (1) by equation (2) gives
PAPB\frac{P_A}{P_B} = mAmBMBMA\frac{m_A}{m_B} \frac{M_B}{M_A}
PA+PBP_A + P_B = 3 ⇒ PBP_B = 3 - 2 = 1 bar
MAMB\frac{M_A}{M_B} = (mAmB)(PBPA)\left(\frac{m_A}{m_B}\right)\left(\frac{P_B}{P_A}\right) = (1 g2 g)(1 bar2 bar)\left(\frac{1\,\text{g}}{2\,\text{g}}\right)\left(\frac{1\,\text{bar}}{2\,\text{bar}}\right) ⇒ MAMB\frac{M_A}{M_B} = 1 : 4
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