1 liter of an ideal gas `(gamma =1.5)` at `300k` is suddenly compressed to half its original volume. (a) Find the ratio of the final pressure to the initial pressure. (b) If the original pressure is `100 kpa`, find the work done by the gas in the process. (c) What is the change in internal energy? (d) What is the final temperature? (e) the gas is now colled to `300k` keeping its pressure constant. Calculate the work done during the process . (f) The gas is now expanded iasothermally to achive its original volume of 1 liters. Calculate the work done by the gas .(g) Calculate the total work done in the cycle.

Explanation

`gamma=1.5`,`T=300K`,`V_i=1L`,`V_2=1/2L`
(a) The process is adiabatic because volume is suddenly change
`P_1V_1^gamma = P_2V_2^gamma`
or `P_2 = P_1((V_1)/(V_2))^gamma=P_1(2)^gamma`
or `P_2/P_1 = 2^1.5 = 2 sqrt2`
(b) `P_1 =100 KPa = 10^5 Pa`
and `P_2 = 2 sqrt2 xx 10^5 Pa`
Work done by adiabatic process
`=(P_1V_1 - P_2V_2)/(gamma-1)`
`=( 10^5 xx 10^(-3) - 2 sqrt2 xx 10^5 xx 1/2 xx 10^(-3))/(15-1)`
`=-82J`
(c) Internal energy,
`dQ=0`
implies ` dU=-dW =-(-82J) =82J`
(d) `T_1V_1^(gamma-1) = T_2V_2^(gamma-1)`
`T_2 = T_1((V_1)/(V_2))^(gamma-1)`
`300(2)^0.5`
`=300 xx sqrt2 xx =300 xx 1.414`
`=T_2 = 424K`
(e) The pressure is kept constant. The process is isobaric work done `= nRdT`,
Here , `n=(PV)/(RT) = (10^5 xx 10 ^(-3))/(R xx 300) = (1)/ (3R)`
So work done `= (1)/(3R) xx R(300-424) `
`=-41.4J`.
(f) `V_1/T_1 = V_2/T_2` ...(1)
`V_1 = V_2(T_1)/(T_2)`
Work done in this process
`=nRTl n (V_1)/(V_2)`
`(1)/(3R) xx R xx T xx In2`
` =100 xx l n 2 xx = 100 xx 1.039`
`=103`
(g) Net work done
`=-82-41.4 +103`
`=-20.4J`