What is the pressure of 2 mole of `NH_(3)` at `27^(@)C` when its volume is 5 lit. in can der Waal's equation ? `(a = 4.17, b = 0.03711)`

Options

A. 10.33 atm
B. 9.333 atm
C. 9.74 atm
D. 9.2 atm

Answer

B

Explanation

`T = 27 + 273 = 300 K`,
`R = 0.0821 L atm K^(-1) mol^(-1)`
`n = 2 mol, " " a = 4.17`
`V = 5L, " " b = 0.03711`
`(P + (an^(2))/(V^(2))) (V - nb) = nRT`
`(P + (4.17 xx 2^(2))/(5^(2))) (5 - 2 xx 0.3711) = 2 xx 0.0821 xx 300`
`(P + (4.1 xx 4)/(25)) (5 - 0.07422) = 600 xx 0.0821`
`(P + 0.67) (4.93) = 600 xx 0.0821`
`P + 0.67 = (600 xx 0.0821)/(4.93)`
`P + 0.67 = 9.99`
`P = 9.99 - 0.67 = 9.32 atm`