A gaseous mixture enclosed in a vessel of volume V consists of one gram mole if a gas ‘A’ with gamma = (C_p/C_v) = 5/3and another gas .B. withgamma = 7/5at a certainof gases A and B are 4 and 32 respectively. The gases do not react with each other and are assumed to be ideal. The gaseous mixture follows the relation PV^(19//13)= constant in an adiabatic process [5^(19//13) = 10.5]The mixture is compressed adiabatically to 1/5 of its volume. Find the change in its adiabaticcompressibility in terms of V and T

A gaseous mixture enclosed in a vessel of volume V consists of one gra

1.	A gaseous mixture enclosed in a vessel of volume V consists of one gram mole if a gas ‘A’ with gamma = (C_p/C_v) = 5/3and another gas .B. withgamma = 7/5at a certainof gases A and B are 4 and 32 respectively. The gases do not react with each other and are assumed to be ideal. The gaseous mixture follows the relation PV^(19//13)= constant in an adiabatic process [5^(19//13) = 10.5]The mixture is compressed adiabatically to 1/5 of its volume. Find the change in its adiabaticcompressibility in terms of V and T
Answer» `0.0248 V/T` `0.426 V/T` `0.0426V/T` `0.684V/T` ANSWER :A