Question

Question: Air that initially occupies 0.140 m³at a gauge pressure of 103 kPais expanded isothermally to a pressure of 101.3 kPaand then cooled at constant pressure until it reaches its initial volume.

Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure).

Short answer

Answer

Work done by the air is$5.60\times {10}^3\text{J}.$

Step-by-step solution

Step 1: Given

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

Determining the concept and formulas

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.

Determine the work done by the air

Initial pressure of the gas is,

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

Work done by an ideal gas during isothermal process is,

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

But,

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

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:

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

Total work done by the air is,

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

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

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