A certain volume of dry air at `20^(@)C` is expanded to three thimes its volume (i) Slowly, (ii) suddenly. Calculate the final pressure and temperature in each case. Atmosperic pressure `= 10^(5) Nm^(-2), gamma` of air = 1.4

Answer

`2.15xx10^(4N),^(-2)`

Explanation

When the process is slow, temperature reamins constant and it maty be taken as an isothermal process for which
pV=a cosntant
`therefore 10^(5)V=p_(2)3 V or p_(2)=(10^(5)/(3)=3.3xx10^(4)Nm^(-2)`
When the process is sudden, it adiabatic
In a adiabatic change ltbRgt `TV^(gamma-1)= a constant`
or `293V^(0.4)=(273+t)3^(0.4),or(273+t)=(293)/(3)^(0.4)`
`therefore log(273+t)=log293-log3^(0.4)=log293-0.4log3`
`therefore 273+1 = antilog 2.2761 = 188.8`
For an adiabatic change `pV^(gamma)=constant`
`therefore 10^(5)V^(1.4)=p_(2)(3V)^(1.4)`, or `10^(5)V^(1.4)V^(1.4)`
or `p_(2)=106(5)/(3)6(1.4)`
`therefore logp_(2)-log 106(5)-1.4log3=5-1.4xx0.4771=5-0.6679=4.3321`
`therefore P_(2) = antilog 4.3321 = 2.15xx10^(4)Nm^(-2)`
In the slow process final temperature `= 20^(@)C`
and pressure `= 3.3xx106(4)Nm^(-2)`
In the sudden process final temperature`=-84.2^(@)C`
and pressure `= 3.3 xx10^(4)Nm^(-2)`
In the sudden process final tempeature`=-84.26^(@)C`
and pressure `= 2.15xx10^(4)Nm^(-2)`