Interesting problem on Pressure!
A container contains mixture of 20 moles of gases is at 50^{0}C (degree Celsius) have volume 22 L (litre) .If 5gm of gases exert pressure on the container is P {1} and it's vapour density (VD) is 50 ,then find P {1} in atm (rounded upto two decimal)
The answer is 0.06.
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1 solution
we know that, t(temperature) = 50^{0}C , n(number of moles )=20 moles , V(volume )=22 L , W (weight )=5gm ,VD=50 Molecular weight = 2 * VD. MW = 2 * 50=100. therefore GMW = 100gm. number of moles (n {1}) =weight of gas/GMW of gas. n {1} = 5gm/100gm............... 1 n {1} =0.05. T = 273+ t. = 273+50. T = 323 K. Pressure (P)= nRT/V. P = 20 (mol)* 0.0821 litre atm/mol K*323K / 22 litre (R=0.0821 litre
atm/mol K). P = 24.10755 atm...........2. now , P {1} (partial pressure ) = n {1}/n (mole fraction) * P. P {1} = 0.05 mole /20 moles * 24.10755 atm(from 1&2). P {1} = 0.0025*24.10755 atm. P {1} = 0.06027 atm. Therefore P_{1} = 0.06 atm