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InterviewSolution
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A vessel of volume , `V = 5.0` litre contains `1.4 g` of nitrogen at a temperature `T = 1800 K`. Find the pressure of the gas if `30%` of its molecules are dissociated into atoms into atmos at this temperature. |
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Answer» The vessel contains a mixture of molecular nitrogen and atomic nitrogen . Let `m_(1), m_(2)` be the masses of teh molecular nitrogen and atomic nitrogen respectively. Then `m_(1) = 1.4 xx 70/100 = 0.98 g , m_(2) = 1.4 xx 30/100 = 0.42 g` Molecular wt. of nitrogen , `M_(1) = 28 , T= 1800 K, V= 5` litre `= 5 xx 10^(-3)m^(3)` Atomic wt. of nitrogen, `M_(2) = 14, R = 8.3 j "mole"^(-1)K^(-1)` Let `P_(1)` and `P_(2)` be the pressure due to molecular and atomic nitrogen respectively. therefore, `P_(1) = (mu_(1)RT)/(V) = (m_(1)RT)/(M_(1)V) = (0.98)/(28) xx (8.3 xx 1800)/(5 xx 10^(-3)) = 1.046 xx 10^(5) Nm^(-2)` `P_(2) (mu_(2)RT)/(V) = (m_(2))RT)/(M_(2)V) = 0.42/14 xx (8.3 xx 1800)/(5 xx 10^(-3)) = 0.894 xx 10^(5) Nm^(-2)` `:.` Net pressure of the gas = `P_(1) +P_(2) = 1.046 xx 10^(5) +0.894 xx 10^(5) = 1.94 xx 10^(5) Nm^(-2)` . |
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