91Ó°ÊÓ

Number of moles n = 2 , Temperature T = 65C = 338K , and the pressure triples from to

Here the is compressed in a cylinder, therefore the work done is negative.

We, have already seen that the pressure is tripled from$P_2$which is equals to$3P_1$. Hence the volume will also decrease three times from$V_1$to$V_2$.

Therefore, the pV diagram of the system will be:

Now, When the pressure changes the work done is given by: -6180J…………(i)

From, the Ideal gas equation we get: $P=\frac{nRT}{V}$

Now, substituting the value of P in the equation (i) we get:

$W=nRT{∫}_{V_1}^{V_2}\frac{1}{V}dV$, here nRT is constant.

$W=nRT\ln \frac{V_2}{V_1}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(¾±¾±)

Again, from Ideal Gas law we obtain: role="math" localid="1664296645122" $\frac{V_2}{V_1}=\frac{P_1}{P_2}$, where $P_2=3P_1$.

Now, substituting the value of$\frac{V_2}{V_1}$ in equation (ii), and calculate the work done .

$$\begin{gathered} W=nRT\ln \frac{P_1}{P_2} \\ W=nRT\ln \frac{P_1}{3P_2} \\ =2mol×8.314J/mol.K338K\ln \frac{1}{3} \\ =-6180J \end{gathered}$$

Therefore, the work done by the system is -6180J