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

 p1 = 0.1 MPa = 105 Pa initial pressure of the gas T1 = 293 K initial temperature of the gas V1 = 830 l = 0.83 m3 initial volume of the gas W = 166.8 kJ = 166.8·103 J work performed during compression of the gas κ = 1.25 Poisson's ratio p2 = ? pressure of the gas after compression V2 = ? volume of the gas after compression T2 = ? 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 V2

Final volume of the gas V2 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 p2

To calculate the final pressure p2 of the gas, use the above-mentioned Poisson's equation.

• Hint 4: Final Temperature T2

To calculate the final temperature T2 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 p pressure during compression is not constant, the following formula for the performed work holds true

$W = -\int\limits_{V_1}^{V_2}p \, \text{d}V,$

where V1 and V2 are initial and final volume of the gas respectively.

To express pressure p as a function of volume V, we use the Poisson's equation for adiabatic processes:

$p_1V_{1}^{\kappa} = pV^{\kappa},$

where κ is the Poisson's ratio. After expressing pressure p from this equation, we obtain

$p = \frac{p_1V_1^{\kappa}}{V^{\kappa}}.$

Now 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 final volume of the gas V2:

$\left(\frac{V_2}{V_1}\right)^{1-\kappa}= \frac{(\kappa-1)W} {p_1V_1}+1,$

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 resultant pressure of the gas p2:

$p_2=\frac{p_1V_1^{\kappa}}{V_2^{\kappa}}.$

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 resultant temperature of the gas T2 after compression:

$T_2=\frac{p_2V_2T_1}{p_1V_1}.$
• 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 m3 and 440 K respectively.