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.

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`.