## Pressure, Volume and Temperature of a Compressed Gas

### Task number: 2180

An ideal gas occupies a volume of 830 litres at a pressure of 0.1 MPa and temperature of 293 K. We compress the gas and perform a work of 166.8 kJ. Determine the resultant pressure, volume and temperature of the gas after compression. Poisson's ratio for the described process taking place in a thermally isolated system is *κ* = 1.25.

#### Notation

*p*_{1}= 0.1 MPa = 10^{5}Painitial pressure of the gas *T*_{1}= 293 Kinitial temperature of the gas *V*_{1}= 830 l = 0.83 m^{3}initial volume of the gas *W*= 166.8 kJ = 166.8·10^{3}Jwork performed during compression of the gas *κ*= 1.25Poisson's ratio *p*_{2}= ?pressure of the gas after compression *V*_{2}= ?volume of the gas after compression *T*_{2}= ?temperature of the gas after compression #### Hint 1

As it is said in the task assignment, the process takes place in a thermally isolated system. What does it mean?

#### Hint 2: Determining Volume

*V*_{2}Final volume of the gas

*V*_{2}can be determined from the formula for the performed work.Think about what this formula looks like when pressure of the gas is not constant (but is the function of volume).

#### Hint 3: Final Pressure

*p*_{2}To calculate the final pressure

*p*_{2}of the gas, use the above-mentioned Poisson's equation.#### Hint 4: Final Temperature

*T*_{2}To calculate the final temperature

*T*_{2}of the gas, use the ideal gas law.#### Analysis

To determine the

**final volume**of the gas, we use the formula for the performed work. Since pressure is a function of volume, we will have to use the integral form of this formula. Thus, we express pressure from the Poisson's equation first and then substitute it in the integral for the obtained function. Once integrated, we can express the final volume.After that we use the Poisson's equation again to determine the

**final pressure**of the gas.**Final temperature**of the gas after compression can be determined from the ideal gas law.#### Solution

Since pressure of the gas

\[W = -\int\limits_{V_1}^{V_2}p \, \text{d}V,\]*p*pressure during compression is not constant, the following formula for the performed work holds truewhere

*V*_{1}and*V*_{2}are initial and final volume of the gas respectively.To express pressure

\[p_1V_{1}^{\kappa} = pV^{\kappa}, \]*p*as a function of volume*V*, we use the Poisson's equation for adiabatic processes:where

\[p = \frac{p_1V_1^{\kappa}}{V^{\kappa}}.\]*κ*is the Poisson's ratio. After expressing pressure*p*from this equation, we obtainNow we substitute pressure in the formula for the work

\[W = -\int\limits_{V_1}^{V_2}p \, \text{d}V = -\int\limits_{V_1}^{V_2}\frac{p_1V_1^{\kappa}}{V^{\kappa}} \, \text{d}V =\]factor out the constants and pull them out of the integral

\[= -p_1V_1^{\kappa} \int\limits_{V_1}^{V_2}\frac{1}{V^{\kappa}} \, \text{d}V =\]integrate and substitute for the limits

\[= -p_1V_1^{\kappa}\frac{1}{-\kappa + 1}\left[V^{-\kappa + 1}\right]_{V_1}^{V_2}= \frac{p_1V_1^{\kappa}}{\kappa-1 }\left(V_2^{-\kappa+1} - V_1^{-\kappa+1}\right).\]Therefore, work performed during the gas compression is

\[W= \frac{p_1V_1^{\kappa}}{\kappa-1}\left(V_2^{1-\kappa} - V_1^{1-\kappa}\right) = \frac{p_1V_1}{\kappa-1}\left[\left(\frac{V_2}{V_1}\right)^{1-\kappa} - 1\right] .\]Therefrom we can express the

\[\left(\frac{V_2}{V_1}\right)^{1-\kappa}= \frac{(\kappa-1)W} {p_1V_1}+1,\]**final volume**of the gas*V*_{2}:and so:

\[V_2=V_1 \sqrt[1-\kappa]{\frac{(\kappa-1)W}{p_1V_1}+1}.\]Then, using the Poisson's equation again

\[p_1V_1^{\kappa}=p_2V_2^{\kappa} \]we can easily determine the

\[p_2=\frac{p_1V_1^{\kappa}}{V_2^{\kappa}}.\]**resultant pressure**of the gas*p*_{2}:Finally, we use the following form of the ideal gas law

\[\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2},\]from which we express the

\[T_2=\frac{p_2V_2T_1}{p_1V_1}.\]**resultant temperature**of the gas*T*_{2}after compression:#### Numerical Substitution

\[V_2=V_1\sqrt[1-\kappa]{\frac{(\kappa-1)W}{p_1V_1}+1}\] \[V_2=0.83\cdot\sqrt[1-1.25]{\frac{(1.25-1)\cdot166.8\cdot{10^3}}{10^5\cdot{0.83}}+1}\,\mathrm{m^3}\dot{=}0.163\,\mathrm{m^3}\] \[p_2=\frac{p_1V_1^{\kappa}}{V_2^{\kappa}}=\frac{10^5\cdot{0.83^{1.25}}}{0.163^{1.25}}\,\mathrm{Pa}\dot{=}765000\,\mathrm{Pa}=765\,\mathrm{kPa}\] \[T_2=\frac{p_2V_2T_1}{p_1V_1}=\frac{765000\cdot{ 0.163}\cdot{293}}{10^5\cdot{ 0.83}}\,\mathrm{K}\dot{=}440\,\mathrm{K}\]#### Answer

The resultant pressure, volume and temperature of the compressed gas are approximately 765 kPa, 0.163 m

^{3}and 440 K respectively.