1.

The average speed at T_(1)Kand the most probable speed at T_(2) K of CO_(2) gas is 9 xx 10^(4) cm s^(-1). Calculatethe value of T_(1) and T_(2).

Answer»

Solution :We know, for 1 MOLE of an IDEAL gas,
rms speed `= SQRT((3RT)/(M))`
and rms speed : average speed : most probable speed
`=1 : 0.9211 : 0.8165`.
`therefore` average speed at `T_(1)K = 0.9211 xx sqrt((3RT_(1))/(M)) = 9 xx 10^(4)` …..(1)
and most probable speed at `T_(2)K = 0.8165 xx sqrt((3RT_(2))/(M)) = 9 xx 10^(4)`.
Substituting `R = 8.314 xx 10^(7)` ergsK/mole and M = 44 in (1) and (2), we get,
`T_(1) = 1684` and
`T_(2) = 2143 K`.


Discussion

No Comment Found

Related InterviewSolutions