A given sample of an ideal gas `(gamma = 1.5 )` is compressed adiabatically from a volume of `150 cm ^(3)` to `50 cm ^(3)`. The initial pressure and the initial temperature are `150 kpa` and `300K`. Find (a) the number of moles of the gas in the sample, (b) the molar heat capacity at constant volume, (c) the final pressure and temperature, (d) the work done by the gas in the process and (e) the change in internal energy of the gas .

Explanation

`PV=nRT`
Given, `P=150 KPa =150 xx 10^3Pa`, `V= 150 cm^3 = 150 xx 10 ^(-6)m^3`
`T=300K`
(a `n=(PV)/(RT)=9.036 xx 10^(-3)`
`=0.009` moles.
(b) `(C_p)/(C_v) = gamma`, `C_p-C_v = R`
So, `C_v = (R)/(gamma-1)=8.3/0.5 = 16.65//mol es`.
(c) Given ,
`P_1 = 150 KPa = 150 xx 10^3 Pa`,
`P_2 = ? V_1 = 150 cm^3`
`=150 xx 10^(-6)m^3`
`gamma=1.5`
`V_2 = 50 cm^3 = 50 xx 10^(-6) m^3`,
`T_1 = 300K`, `T_2=?`
Since the process is adiabatic hence-
`P_1V_1^gamma = P_2V_2^gamma`
implies `150xx10^3 xx (150 xx 10^(-6))^gamma`
`=P_2 xx ( 50 xx 10^(-6))^gamma`
implies `P_2 = 150 xx 10^3 xx (150xx 10^6)^(1.5)/(50 xx 10 ^(-6))^1.5`
`=150000xx(3)^1.5`
`779.422 xx 10^8 Pa`
`=780 KPa`
Again,
`P_1^(1-gamma) T_1^(gamma) = P_1^(1-gamma)T_2^(gamma)`
implies `(150 xx 10^3)^(1-1.5) xx (330)^1.5`
`=(780 xx 10 ^3) ^(1-1.5) xx T_2^1.5`
implies `T_2^1.5 = (150 xx 10 ^3)^(1-1.5) xx (300)^1.5 xx 300 ^1.5`
`=11849.050`
implies `T_2= (11849.050)^(1//1.5)`
`=519.74 = 520`
(d) `dQ=W + dU`
or ` W=-dU`[`dQ=0`, in adiabatic]
`= -nCvdT`
`=-0.009 xx 16.6 xx (520 -300)`
= `-0.009 xx 16.6 xx 220`
`=-32.8J = - 33J`.
(e) `dU=nCvdT`
`=0.009 xx 16.6 xx 220 = 33J`.