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If the ration of the masses of `SO_(3)` and `O_(2)` gases confined in a vessel is `1 : 1` , then the ratio of their partial pressure would beA. `5 : 2`B. `2 : 5`C. `2 : 1`D. `1 : 2` |
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Answer» Correct Answer - B Let m be the mass of `SO_(3)` and `O_(2)` enclosed in the vessel. Number of moles of `SO_(3)=(m)/(80)` Number of moles of `O_(2)=(m)/(32)` Partial pressure of `SO_(3), P_(A)=(m)/(80)xx(R xx T)/(V)` Partial pressure of `O_(2), P_(B)=(m)/(32)xx(R xx T)/(V)` Now, `(P_(A))/(P_(B))=(m)/(80)xx(32)/(m)=(2)/(5)` Hence, ratio of partial pressure of m g of `SO_(3)` and `O_(2)` is 2 : 5. |
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