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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. |
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Answer» SOLUTION :Suppose the molecular mass of the ideal gas A is MA and that of B is Me. When only the ideal gas A is PRESENT, `PV = nRT` `2xx V = 1/M_Axx RxxT (:. n = m/M =1/M) ...(i)` On mixing the TWO gases, total number of moles `=1/M_A+2/M_B` Hence, according to the gas EQUATION `PV = nRT` `3xxV=(1/M_A+2/M_B)xxRxxT... (ii)` Dividing eq. (ii) by eq. (i), we have `3/2 = ((M_B + 2M_A)/(M_A M_B))/(1/M_A)=(M_B+2M_A)/M_B =(1+2xxM_A)/M_B` or `2xxM_A/M_B = 3/2 - 1 = 1/2` or `M_A/M_B = 1/2 xx 1/2 = 1/4` or `M_B =4xxM_A` |
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