Question

Two moles of an ideal gas are compressed in a cylinder at a constant temperature of 65.0°C until the original pressure has tripled. (a) Sketch a pV-diagram for this process. (b) Calculate the amount of work done.

Short answer

a) The pV diagram for the process is:

b) The amount of work done by two moles of an ideal gas compressed in a cylinder at a constant temperature of$65.0C°$is$-6180J$

Step-by-step solution

Work done by an ideal gas compressed in a cylinder.Given,

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:

Calculating the amount of work done by the system.

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:

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

$W=nRT\ln \frac{V_2}{V_1}$……………………………(ii)

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 .

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

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