Question

When a certain gas under a pressure of 4.90 106 pa at 20.0°c is allowed to expand to 3.00 times its original volume, its final pressure is 1.06 106 pa. What is its final temperature?

Answer 1

As per question the initial states of the gases are given as

INITIAL STATE: FINAL STATE:

`p_(1) =4.90106 pa`
`p_(2) =1.06106 pa`

`v_(1) =v[say]`
`v_(2) =3v[say]`

`T_(1) =20 degree celcius  =293 K`
`T_(2) =?`

AS per combined gas equation obtained from the combination of Boyle's law and Charles law [Basic ideal gas laws]

`(p_(1) v_(1) )/(T_(1) ) =(p_(2)v_(2)  )/(T_(2) )`

Hence
`T_(2) =(p_(2) v_(2)T_(1)  )/(p_(1) v_(1) )`

=
`(1.06106*3v*293)/(4.90106*v)`

=190.3 K [ANS]