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]

              $\frac{p_{1} v_{1} }{T_{1} } =\frac{p_{2}v_{2}  }{T_{2} }$

Hence $T_{2} =\frac{p_{2} v_{2}T_{1}  }{p_{1} v_{1} }$

                    =$\frac{1.06106*3v*293}{4.90106*v}$

                    =190.3 K [ANS]