Is the gas heated or cooled?

1. Jul 19, 2016

Vibhor

1. The problem statement, all variables and given/known data

2. Relevant equations

PV = nRT

3. The attempt at a solution

$PV^2 = constant$

Differentiating both the sides we get , $V^2dP + 2PVdV = 0$

Now , using this along with $PdV+VdP=nRdT$ , we get $PdV = - nRdT$ .

Since $dV$ is positive , $dT$ is negative which means the temperature falls or gas is cooled .

Is this the correct way to analyze the problem ?

Thanks

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2. Jul 19, 2016

Staff: Mentor

Yes.

3. Jul 19, 2016

Vibhor

Thank you .

Please check the attached image containing the official solution . Do you think the analysis done in it is correct ? I could not understand it .

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4. Jul 19, 2016

Staff: Mentor

For the same fractional increase in volume the final pressure is lower, so the temperature must have decreased rather than staying constant. It's equivalent to what you did mathematically, but I like the way you did it better.

5. Jul 20, 2016

andrevdh

A gas does positive work when it expands.
The internal energy is used for this resulting in a decrease in the gas's temperature, unless heat is supplied to cancel this.
When just enough heat is supplied to cancel this we get that Boyle's law is upheld, or put another way the gas is kept at a constant temperature by a heat source.
This means that the heat into the system is used to do the work or produce the expansion.
In this case the gas has to expand even more to keep the product constant (V2).

6. Jul 20, 2016

Staff: Mentor

There is no need to invoke internal energy or the first law to address this problem. This can be done strictly as an ideal gas equation analysis.

7. Jul 20, 2016

Vibhor

8. Jul 20, 2016

Staff: Mentor

I saw it, but I wasn't able to figure out exactly what they are asking.

9. Jul 20, 2016

Vibhor

Ok

10. Jul 20, 2016

TSny

Another approach is to substitute $P = nRT/V$ into $PV^2 =$constant to get $TV =$ constant.

11. Jul 20, 2016

Vibhor

Fantastic ! You made my work in OP look silly .