Question

When 53000000J of heat is added to a gas enclosed in a cylinder fitted with a light friction less piston maintained at atmospheric pressure,the volume is observed to increase from 1.9m^3 to 4.1 m^3.What is the change in internal energy of the gas?

Answer 1

Clearly,on addition of heat,only volume change took place,but the pressure of the gas due to the weight of the piston and atmospheric pressure remained the same.

So,this was an isobaric process.

Now,form `1` st law of thermodynamics,

`delQ= delU +delW=delU +P delV`

Now,as the piston is very light,we can omit the pressure exerted by it,so here, `P=10^5 Pa` i.e atmospheric presure

Given, `delQ=53*10^6 J` and, `delV=(4.1-1.9)=2.2 m^3`

So,putting the values we get,change in internal energy=

`del U=53*10^6-2.2*10^5=530*10^5 -2.2*10^5=527.8*10^5 J`