91Ó°ÊÓ

$$\begin{gathered} V_i=0.140{\text{m}}^3 \\ {P}_{ig}=103.0\text{kPa}=1.03×{10}^5\text{Pa} \\ {P}_f=101.3\text{kPa}=1.013×{10}^5\text{Pa} \end{gathered}$$

Find the work done during isothermal and isobaric processes using corresponding formulae. Adding them will give the work done by the air.

Formula is as follow:

$p_i{v}_i=nR{T}_i$

Here, p is pressure, v is volume, T is temperature, R is universal gas constant and n is number of moles.

Initial pressure of the gas is,

$$\begin{gathered} P_i=P_f+P_{ig} \\ {P}_i=1.013×{10}^5+1.03×{10}^5 \\ {P}_i=2.04×{10}^5\text{Pa} \end{gathered}$$

Work done by an ideal gas during isothermal process is,

$W_T=nRT\ln \left(\frac{P_i}{P_f}\right)$

But,

$$\begin{gathered} P_i{V}_i=nRT \\ {W}_T=P_i{V}_i\ln \left(\frac{P_i}{P_f}\right) \\ {W}_T=\left(2.04×{10}^5\right)\left(0.140\right)\ln \left(\frac{2.04×{10}^5}{1.013×{10}^5}\right) \\ ∴W_T=2.00×{10}^4\text{J} \end{gathered}$$

Thefinal pressure and volume attained in the isothermal process is the initial pressure and volume for the isobaric process.

Work done by an ideal gas during isobaric process is,

$$\begin{gathered} W_P=P_f\left(V_i-V_f\right) \\ But \\ {V}_f=\frac{P_i{V}_i}{P_f} \\ {W}_P=P_f\left(V_i-\frac{P_i{V}_i}{P_f}\right) \\ {W}_P=P_f{V}_i-P_i{V}_i \\ {W}_P=(P_f-P_i)V_i \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} W_P=\left(1.013×{10}^5-2.04×{10}^5\right)0.140 \\ {W}_P=-1.44×{10}^4\text{J} \end{gathered}$$

Total work done by the air is,

$$\begin{gathered} W=W_T+W_P \\ W=2.00×{10}^4+\left(-1.44×{10}^4\right) \\ W=0.56×{10}^4=5.60×{10}^3\text{J} \end{gathered}$$

Hence, the work done by the air is$5.60×{10}^3\text{J}.$

Therefore,total work done by the air during isothermal and isobaric process can be calculated from its initial and final pressures and volumes.