One mole of an ideal monoatomic gas is subjected to changes as shown in the graph. The magnitude of the work done (by the system or on the system) is _____ J (nearest integer)

Given: log 2 = 0.3, ln 10 = 2.3

The cycle consists of three processes:

• $$1 \rightarrow 2$$ : Isobaric expansion
• $$2 \rightarrow 3$$ : Isochoric process
• $$3 \rightarrow 1$$ : Isothermal compression

First, verify that the process $$3 \rightarrow 1$$ is isothermal.

At point 1,

$$P_1V_1=(1.0)(20)=20\ \text{bar·L}$$

At point 3,

$$P_3V_3=(0.5)(40)=20\ \text{bar·L}$$

Since

$$P_1V_1=P_3V_3$$

the temperature remains constant and the process $$3 \rightarrow 1$$ is isothermal.

For the process $$1 \rightarrow 2$$,

$$W_{12}=P\Delta V$$

$$W_{12}=1.0(40-20)$$

$$W_{12}=20\ \text{bar·L}$$

For the process $$2 \rightarrow 3$$,

$$\Delta V=0$$

Therefore,

$$W_{23}=0$$

For the isothermal process $$3 \rightarrow 1$$,

$$W_{31}=P_iV_i\ln\left(\frac{V_f}{V_i}\right)$$

$$W_{31}=20\ln\left(\frac{20}{40}\right)$$

$$W_{31}=20\ln\left(\frac{1}{2}\right)$$

$$W_{31}=-20\ln 2\ \text{bar·L}$$

Hence, the net work done in the cycle is

$$W_{\text{net}}=W_{12}+W_{23}+W_{31}$$

$$W_{\text{net}}=20+0-20\ln 2$$

$$W_{\text{net}}=20(1-\ln 2)\ \text{bar·L}$$

Using

$$\log_{10}2=0.3$$

and

$$\ln 10=2.3$$

$$\ln 2=\ln 10\times \log_{10}2$$

$$\ln 2=2.3\times 0.3=0.69$$

Substituting,

$$W_{\text{net}}=20(1-0.69)$$

$$W_{\text{net}}=20(0.31)$$

$$W_{\text{net}}=6.2\ \text{bar·L}$$

Using

$$1\ \text{bar·L}=100\ \text{J}$$

$$W_{\text{net}}=6.2\times 100$$

$$W_{\text{net}}=620\ \text{J}$$

Therefore, the magnitude of the work done is

$$620\ \text{J}$$