At constant P of 0.0821 atm lo g V vs lo g T graph was plotted for three sample of ideal gases (1), (2) and (3), each having a definite number of moles. If the mixture of three gases is taken in a con

Question

At constant P of 0.0821 atm lo g V vs lo g T graph was plotted for three sample of ideal gases (1), (2) and (3), each having a definite number of moles. If the mixture of three gases is taken in a container of volume 4 L with each gas as same number of moles as they were taken in plotting the graph then what is the partial pressure of Gas (1), in atm, if the total pressure is 9atm? (Take \displaylines lo g 10 ​ 2 = 0.3010 lo g 10 ​ 3 = 0.4771 )

TestHub · Accepted Answer

V ∝ T T V ​ = K = P nR ​ lo g V = lo g T + lo g K y = mx + c C = lo g K = lo g ( P nR ​ ) C = lo g [ n × 0.0821 0.0821 ​ ] For gas ( 1 ) : C = lo g n 0.4771 = lo g n = lo g 3 n = 3 Forgas (2): 0 = lo g n = lo g 1 n = 1 For gas ( 3 ) − lo g x = 0.30 − lo g 2 = lo g n − lo g 2 − 1 = lo g n lo g [ 2 1 ​ ] = lo g n n = 2 1 ​ mole Partial pressure of gas (1) = ( 4.5 3 ​ ) P t o t a l ​ = 6 a t m

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JEE Advanced (2028)