Question

A cylinder with a frictionless, movable piston like that shown in Fig. 19.5 contains a quantity of helium gas. Initially the gas is$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{Pa}$ at 300 K and and occupies a volume of 1.50 L. The gas then undergoes two processes. In the first, the gas is heated and the piston is allowed to move to keep the temperature at . This continues until the pressure reaches $\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{Pa}$. In the second process, the gas is compressed at constant pressure until it returns to its original volume of 1.50 L. Assume that the gas may be treated as ideal. (a) In a pV-diagram, show both processes. (b) Find the volume of the gas at the end of the first process, and the pressure and temperature at the end of the second process. (c) Find the total work done by the gas during both processes. (d) What would you have to do to the gas to return it to its original pressure and temperature?

Short answer

(a) The required diagram is shown.

(b) The volume of the gas at the end of first process is. The pressure and temperature at the end of second process are,

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

(c) The total work done by the gas during both processes is 95J .

(d) By heating the gas maintained at constant volume, the work done will be zero and the pressure

Step-by-step solution

Ideal gas:

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

Identification of the given data.

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

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

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

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

(a) Depiction of the p-V diagram.

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 $1\to 2$is the isothermal process according to the above equation.

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

 (b) Determine the volume of the gas at the end of the first process and the pressure and temperature at the end of the second process.

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,

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

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,

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

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

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

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

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

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

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

 (c) Determine the total work done by the gas during both processes.

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 $1.00\times {10}^5\mathrm{Pa}$ for ${\mathrm{P}}_1$, $2.40\times {10}^4\mathrm{Pa}$for ${\mathrm{P}}_2$, and $1.5\times {10}^{-3}\hspace{0.33em}{\mathrm{m}}^3$ for in the above equation.

Thus, the work done in the first process is,

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

For the second process, the work done is,

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

Substitute all the values mentioned above,

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

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

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

Thus, the total work done is 95 J.

(d) Determine the process to make the gas return to its original pressure and temperature.

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.