Saved Bookmarks
| 1. |
A gaseous mixture enclosed in a vessel of volume V consists of one mole of a gas A with `gamma=(C_p//C_v)=5//3` and another gas B with `gamma=7//5` at a certain temperature T. The relative molar masses of the gasses A and B are 4 and 32, respectively. The gases A and B do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation `PV^(19//13)=constant`, in adiabatic processes. (a) Find the number of moles of the gas B in the gaseous mixture. (b) Compute the speed of sound in the gaseous mixture at `T=300K`. (c) If T is raised by 1K from 300K, find the `%` change in the speed of sound in the gaseous mixture. (d) The mixtrue is compressed adiabatically to `1//5` of its initial volume V. Find the change in its adaibatic compressibility in terms of the given quantities. |
|
Answer» (a) As for ideal gas `C_(P) - C_(V) = R`and `gamma = (C_(P)//C_(V))`, So, `gamma-1 =(R )/(C_(V))` or `C_(V) = (R )/((gamma-1))` `(C_(V))_(1) = (R )/((5//3)-1) = (3)/(2)R`, `(C_(V))_(2) = (R )/((7//5)-1) = (5)/(2)R` and `(C_(V))_(mix) = (R )/((19//13)-1) = (13)/(6)R` Now from conservation of energy, i.e., `DeltaU = DeltaU_(1) +DeltaU_(2)`, `(n_(1)+n_(2)) (C_(V))_(mix) DeltaT = [n_(1)(C_(V))_(1) +n_(2) (C_(V))_(2)] DeltaT` i.e., `(C_(V))_(mix) = (n_(1)(C_(V))_(1)+n_(2)(C_(V))_(2))/(n_(1)+n_(2))` We have `(13)/(6)R = (1xx(3)/(2)R+nxx(5)/(2)R)/(1+n)=((3+5n))/(2(1+n))` or, `13 + 13n = 9 +15n`, (b) Molecular weight of the mixture will be given by `M = (n_(A)M_(A)+n_(B)M_(B))/(n_(A)+n_(B)) = ((1)(4)+2(32))/(1+2)` `M = 22.67` Speed of sound in a gas is given by `v = sqrt((gamma RT)/(M))` Therefore, in the mixture of the gas `v = sqrt(((19//13)(8.31)(300))/(22.67 xx 10^(-3)))m//s` `v ~~ 401 m//s` (c) `v prop sqrt(T)` or `v = KT^(1//2).....(2)` `rArr (dv)/(dT) = (1)/(2) KT^(-1//2)` `rArr dv = K ((dT)/(2sqrt(T)))` `rArr (dv)/(v) = (k)/(v) ((dT)/(2sqrt(T)))` `rArr (dv)/(v) = (1)/(sqrt(T)) ((dT)/(2sqrt(T))) = (1)/(2) ((dT)/(T))` `rArr (dv)/(v) xx 100 =(1)/(2) ((dT)/(T)) xx 100 = (1)/(2) ((1)/(300)) xx 100 = 0.167 = (1)/(6)` Therefore, percentage change in speed is `0.167%`. |
|