An ideal gas has a specific heat at constant pressure `C_P=(5R)/2`. The gas is kept in a closed vessel of volume `0.0083m^3`, at a temperature of 300K and a pressure of `1.6xx10^6 N//m^2`. An amount of `2.49xx10^4` Joules of heat energy is supplied to the gas. Calculate the final temperature and pressure of the gas.

Explanation

`C_(V) = C_(P)- R = (5R)/(2) - R = (3R)/(2) , DeltaV =0, T_(1) = 300K, V=0.0083 m^(3), P_(1) = 1.6xx10^(6) Nm^(-2)`
From first law of thermodynamics `Q= Delta U+PDeltaV implies DeltaU Q = 2.49 xx 10^(4) J`
From gas equation , `n=(PV)/(RT) = (1.6xx 10^(6) xx 0.0083)/(8.3xx 300) = 16/3`
` therefore DeltaU = nC_(v)DeltaT implies DeltaT = (DeltaU)/(nC_(v)) = (2.49xx10_(4)xx6)/(3xx8.3xx16) = 375K`
Final temperature =`300+375 = 675K`
According to pressure law `PpropT implies (P_(2))/(P_(1)) = (T_(2))/(T_(1)) implies P_(2) = (T_(2))/(T_(1)) xx P_(1) = (1.6xx10^(6) xx 675)/(300) = 3.6xx10^(6)Nm^(-2)`