Question

An ideal gas initially at 4.00atm and 350 K is permitted

toexpand adibatically to 1.50 times its initial volume. Findthe
final pressure and temperature if the gas is a)monoatomic;
b)diatomic with Cv=.

Answer 1

Step-by-step explanation:

It is given that initially pressure of ideal gas is 4.00 atm and its temperature is 350 K. Let us assume that the final pressure is
`P_(2)` and final temperature is
`T_(2)`.

(a) We know that for a monoatomic gas, value of
`\gamma` is \frac{5}{3}[/tex].

And, in case of adiabatic process,

`PV^(\gamma)` = constant

also, PV = nRT

So, here
`T_(1)` = 350 K,
`V_(1) = V`, and
`V_(2) = 1.5 V`

Hence,
`(T_(2))/(T_(1)) = ((V_(1))/(V_(2)))^(\gamma -1)`

`(T_(2))/(350 K) = ((V)/(1.5V))^{(5)/(3) -1}`

`T_(2)` = 267 K

Also,
`P_(1)` = 4.0 atm,
`V_(1) = V`, and
`V_(2) = 1.5 V`

`(P_(2))/(P_(1)) = ((V_(1))/(V_(2)))^(\gamma)`

`(P_(2))/(4.0 atm) = ((V)/(1.5V))^{(5)/(3)}`

`P_(2)` = 2.04 atm

Hence, for monoatomic gas final pressure is 2.04 atm and final temperature is 267 K.

(b) For diatomic gas, value of
`\gamma` is \frac{7}{5}[/tex].

As,
`PV^(\gamma)` = constant

also, PV = nRT

`T_(1)` = 350 K,
`V_(1) = V`, and
`V_(2) = 1.5 V`

`(T_(2))/(T_(1)) = ((V_(1))/(V_(2)))^(\gamma -1)`

`(T_(2))/(350 K) = ((V)/(1.5V))^{(7)/(5) -1}`

`T_(2)` = 289 K

And,
`P_(1)` = 4.0 atm,
`V_(1) = V`, and
`V_(2) = 1.5 V`

`(P_(2))/(P_(1)) = ((V_(1))/(V_(2)))^(\gamma)`

`(P_(2))/(4.0 atm) = ((V)/(1.5V))^{(7)/(5)}`

`P_(2)` = 2.27 atm

Hence, for diatomic gas final pressure is 2.27 atm and final temperature is 289 K.