91影视

A gas that strictly abides by the ideal-gas law: a gas in which there is typically no molecular attraction.

Initial pressure of the gas is ${\mathrm{P}}_1=1.00脳{10}^5\mathrm{Pa}$.

Initial temperature of the gas is${\mathrm{T}}_1=300\mathrm{K}$.

Final pressure of the gas is ${\mathrm{P}}_2=2.50脳{10}^4\mathrm{Pa}$.

Initial volume of the gas is ${\mathrm{V}}_1=1.50\mathrm{L}$.

The required diagram is,

The first process says that the temperature remains constant, i.e. the process is isothermal. The gas being ideal the equation of state is,

PV = RT

The process $1鈫�2$is the isothermal process according to the above equation.

The second process says the pressure remains constant, i.e. the process is isobaric. Process $2鈫�3$shows the isobaric process parallel to the V-axis at constant pressure.

According to the ideal gas equation,

PV = RT

At constant temperature, the product of pressure-temperature is constant as R is the universal gas constant. Thus,

$\begin{array}{r}P=\frac{RT}{V}\\ V鈭�\frac{1}{P}\end{array}$

Solve for the final volume at the end of first process using the known values of the initial pressure, final pressure at the end of first process and initial volume of the gas,

$\begin{array}{r}{V}_2=\left(V_1\right)\left(\frac{P_1}{P_2}\right)\\ =\left(1.50\mathrm{L}\right)\left(\frac{1.00脳{10}^5\mathrm{Pa}}{2.50脳{10}^4\mathrm{Pa}}\right)\\ =6.00\mathrm{L}\end{array}$

The pressure remains constant during the second process so the final pressure at the end of the second process is,

$\begin{array}{r}{\mathrm{P}}_3={\mathrm{P}}_2\\ =2.50脳{10}^4\mathrm{Pa}\end{array}$

The final volume after the compression is same as the initial volume, so solve for the final temperature,

$\begin{array}{r}{T}_3=T_1\left(\frac{P_3}{P_1}\right)\\ =\left(300\mathrm{k}\right)\left(\frac{2.50脳{10}^4\mathrm{Pa}}{1.00脳{10}^5\mathrm{Pa}}\right)\\ =75.0\mathrm{K}\end{array}$

Thus, the volume at the end of first process and the pressure and temperature at the end of second process are,

$\begin{array}{r}{\mathrm{V}}_2=6.00\mathrm{L}\\ {\mathrm{P}}_3=2.50脳{10}^4\mathrm{Pa}\\ {\mathrm{T}}_3=75.0\mathrm{K}\end{array}$

The work done in an isothermal process is,

$\begin{array}{r}W=nRT\ln \left(\frac{V_2}{V_1}\right)\\ =P_1{V}_1\ln \left(\frac{P_1}{P_2}\right)\end{array}$

Substitute $1.00脳{10}^5\mathrm{Pa}$ for ${\mathrm{P}}_1$, $2.40脳{10}^4\mathrm{Pa}$for ${\mathrm{P}}_2$, and $1.5脳{10}^{-3}鈥�{\mathrm{m}}^3$ for in the above equation.

Thus, the work done in the first process is,

$\begin{array}{r}W=\left(1.00脳{10}^5\mathrm{Pa}\right)\left(1.5脳{10}^{鈭�3}{\mathrm{m}}^3\right)\ln 鈦�\left(\frac{1.00脳{10}^5\mathrm{Pa}}{2.40脳{10}^4\mathrm{Pa}}\right)\\ =208\mathrm{J}\end{array}$

For the second process, the work done is,

$\begin{array}{r}\mathrm{W}={\mathrm{P}}_2\left({\mathrm{V}}_3鈭�{\mathrm{V}}_2\right)\\ ={\mathrm{P}}_2\left({\mathrm{V}}_1鈭�{\mathrm{V}}_2\right)\\ ={\mathrm{P}}_2{\mathrm{V}}_1\left(1鈭�\left(\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}\right)\right)\end{array}$

Substitute all the values mentioned above,

$\begin{array}{r}{W}_2=\left(2.40脳{10}^4\mathrm{Pa}\right)\left(1.5脳{10}^{鈭�3}{\mathrm{m}}^3\right)\left(1鈭�\frac{1.00脳{10}^5\mathrm{Pa}}{2.40脳{10}^4\mathrm{Pa}}\right)\\ =鈭�113\mathrm{J}\end{array}$

Thus, the total work done in the entire process is,

$\begin{array}{r}W=W_1+W_2\\ =208\mathrm{J}鈭�113\mathrm{J}\\ =95\mathrm{J}\end{array}$

Thus, the total work done is 95 J.

By heating the gas maintained at constant volume, the work done will be zero and the pressure along with the temperature will be reverted back to their original values.