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balu736 [363]
2 years ago
9

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 p

ressure is 1.06 106 pa. What is its final temperature?
Physics
1 answer:
Elina [12.6K]2 years ago
6 0

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]

                 

   

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