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, B, C, D

Explanation

(a) `(5+7n_B)/(3+5n_B)=19/13 rArr n_b=2mol.`
We know that
`(n_A+n_b)/(gamma_m-1)=(n_A)/(gamma_A-1)+(n_B)/(gamma_B-1)`
Where `gamma_m`=Ratio of specific heats of mixture
Here, `n_A=1, gamma_A=5//3,gamma_B=7//5`
According to the relationship
`PV^(19/13)=constant, we get gamma_m=19/13`
(b) On substituting the values we get `n_B=2mol.`
We know that velocity of sound in air is given by the
relationship
`v=sqrt((gammaP)/d) where d=density =m/V`
Also, `PV=(n_A+n_B)RT rArr PV=(n_A+n_B)/VRT`
`:. v=sqrt((gamma(n_A+n_B)RT)/(Vxxm/V))=sqrt((gamma(n_A+n_B)RT)/m)`
Mass of the gas, `m=n_AM_A+n_BM_B=1xx4+2xx32`
`=68g//mol=0.068kg//mol`
`:. v=((19(1+2)xx8.314xx300)/(13xx0.0608))=400.03 ms^-1`
(c) Velocity of sound,
`v=sqrt((gammaRT)/M) and v+Deltav=sqrt((gammaR(T+DeltaT))/M)`
`rArr(v+Deltav)/v=sqrt((T+DeltaT)/T)=(1+(DeltaT)/T)^(1//2)`
When `DeltaTlt lt T then (DeltaT)/Tlt lt1`
`:. 1+(Deltav)/v=1+1/2 (DeltaT)/T`
Percentage change `(Deltav)/vxx100=1/2xx(DeltaT)/Txx100`
`(Deltav)/vxx100=1/2 1/300xx100=1/6%`
(d) `PV^(gamma)=Const.`
Differentiating thr above equation
`V^(gamma)(dP)-P(gammaV^(gamma-1)dV)=0`
`rArr V^(gamma)(dP)=gammaPV^(gamma-1)dV`
`rArr(dP)/(dV)=(gammaPV^(gamma-1))/(V^(gamma))=gammaPV^(gamma-1-1)=(gammaP)/V`
`rArr (-dP)/(dV//V)=-gammaP`
`:. Bulk Modulus B=gammaP`
`:. Compressibility K=1/B=1/(gammaP)`
`:. K_1=1/(gammaP_1) and K_2=1/(gammaP_2)`
`DeltaK=K_2-K_1=1/(gammaP_2)-1/(gammaP_1)=1/gamma(1/(P_2)-1/(P_1))`
Since the process is adiabatic, `P_2V_2^(gamma)=P_1V_1^(gamma)`
`:. P_2=P_1((V_1)/(V_2))^(gamma)=P_1((V_1)/(V_1//5))^(gamma)=P_15^(gamma)`
`:. DeltaK=1/gamma(1/(P_15^(gamma))-1/(P_1))=1/(gammaP_1)(1/(5^(gamma))-1)`
`P_1=((n_A+n_B)RT)/V=((1+2)xx8.31xxT)/V=(24.93T)/V`
`rArr DeltaK=1/(19/13xx24.93xxT/V)(1/(5^(19//13))-1)`
`=-8.27xx10^-5 V [:' T=300K]`