If the kinetic energy and RMS speed of a gas at a certain temperature are 4 . 0 kJ `mol^(-1) 5 . 0 xx 10^(4) cm s^(-1)` respectivley . The molecular weight of the gas is

Options

A. 16
B. 32
C. 64
D. 44

Answer

B

Explanation

Key Idea First calculate temperature, using the average kinetic energy equation , i.e. KE = `(3)/(2)` RT
The, calculate the molecular weight , usign the formula for root mean square velocity `i.e.v_("rms") = sqrt((3RT)/(M))` .
Given,
Kinetic energy ` = 4.0 kJ mol^(-1)`
`v_("rms") = 5 . 0 xx 10 ^(4) cms ^(-1)`
or `v_("rms") = 5 . 0 xx 10^(2) ms^(-1)` As we know that,
`KE = (3)/(2) RT`
`4 xx 1000J //mol = (3)/(2) xx 8.314 J//mol xx T `
`:. T = (4 xx 1000 xx 2)/(3 xx 8.314) = 320 K`
Now, `v_("rms") = sqrt((3RT)/(M))`
`:. 5 xx 10^(2) m//s = sqrt((3 xx 8.314 J //K //mol xx 320 K)/( M)) " "` (`:. 1 J = 1 kg m^(2) s^(-2)` )
`:. M = (3 xx 8.314 xx 320)/( 25 xx 10^(4)) kg //mol`
` = 0 . 0 3192 kg // mol = 31/.92 ~~ 32 g//mol`
Hence, the molecular weight of the gas in 32 g/mol .