Question

One mole of an ideal gas is heated slowly so that it goes from the PV state (Pi , Vi ) to (3Pi , 3Vi ) in such a way that the pressure is directly proportional to the volume. (a) How
much work is done on the gas in the process? (b)How is the temperature of the gas related to its volume during this process?

Answer 1

Since pressure and volume are directly proportional to one another.
Pi/Pf=Vf/Vi
Pf=(Pi/Vi)*(Vf)
W=-integral of pdv
Set the limits to be 3vi and vi.
I am struggling with the integration part

Answer 2

Let \(p,V\) be the pressure and Volume during the process.
\(\dfrac{p}{V} =\dfrac{p_i}{V_i} \implies p =(\dfrac{p_i}{V_i})*V \)

\[W_{\text{done on gas}} = -\int\limits_{V_i}^{3V_i}pdV = -\int\limits_{V_i}^{3V_i}(\frac{p_i}{V_i})*VdV =-\frac{p_i}{V_i}\int\limits_{V_i}^{3V_i}V\,dV \]

Easy to evaluate that integral right ?

Answer 3

I thought at first that the ratio of Pi to Vi was not constant. Yes,that would be easyto integrate