## Work, Pressure and Heat of the Air during Isothermal Expansion

### Task number: 2179

Ideal gas of a volume of 1 m^{3} at initial pressure of 200 kPa is expanding isothermally to occupy double of its initial volume. Determine the work performed by the gas during expansion, final pressure and the amount of heat supplied to the gas.

#### Notation

*V*_{1}= 1 m^{3}initial volume of the gas *p*_{1}= 200 kPa = 2·10^{5}Painitial pressure of the gas *V*_{2}= 2*V*_{1}volume of the gas after expansion *W*= ?work performed by the gas *p*_{2}= ?final pressure *Q*= ?supplied heat #### Hint 1: Work

*W*Think about how to calculate the work performed by the air if pressure is a function of volume.

#### Hint 2: Expressing Pressure

*p*(*V*)To express pressure

*p*as a function of volume*V*, use the so-called Boyle-Mariotte law.#### Hint 3: Resultant Pressure

To determine the resultant pressure of the air, use the Boyle-Mariotte law again.

#### Hint 4: Supplied Heat

Internal energy

*U*of the gas remains unchanged during an isothermal process, therefore all heat*Q*received by the system will be converted into work*W*performed by the gas.#### Analysis

To calculate the work done by the air, we will have to use integration, because pressure is the function of volume.

First, we express pressure of the air from the so-called Boyle-Mariotte law, which applies to isothermal processes, and integrate the obtained function with respect to volume. Initial and final volume of the air serve as the limits of the integral.

Then we use the Boyle-Mariotte law again to calculate the resultant pressure of the air.

In an isothermal process, heat received by the gas is equal to the work it performs (internal energy does not change).

#### Solution

To the

\[W = \int\limits_{V_1}^{V_2}p(V)\, \text{d}V,\]**work***W*performed by the air during isothermal expansion the general formula can be applied:where

*p*stands for pressure,*V*_{1}and*V*_{2}for initial and final volume of the air respectively.Thus, we need to express pressure

\[p_1V_1 = pV. \]*p*as a function of volume*V*first. To do so, we use the Boyle-Mariotte law, according to which:Therefrom we easily express pressure

\[p = \frac{p_1V_1}{V}.\]*p*:Now we can proceed to the integration itself

\[W = \int\limits_{V_1}^{V_2}p\, \text{d}V = \int\limits_{V_1}^{2V_1}\frac{p_1V_1}{V}\, \text{d}V =\]we factor the constants out of the integral

\[=p_1V_1 \int\limits_{V_1}^{2V_1}\frac{1}{V}\, \text{d}V = \]integrate and substitute the limits

\[=p_1V_1[\ln\,V]_{V_1}^{2V_1} = p_1V_1\,\ln \frac{2V_1}{V_1}= p_1V_1\,\ln\,2.\]The resultant

\[p_1V_1=p_2V_2.\]**pressure***p*_{2}of the air can be determined from the above-mentioned Boyle-Mariotte law as wellTherefrom we express the final pressure

\[p_2=\frac{p_1V_1}{V_2}=\frac{p_1V_1}{2V_1}= \frac{p_1}{2}.\]In an isothermal process, all

\[Q=W.\]**heat***Q*received by the gas is converted into the work*W*that it performs (internal energy*U*does not change). Thus, it holds true:#### Numerical Substitution

\[W=p_1V_1\,\ln\,2= 2\cdot{ 10^5}\cdot 1\cdot \ln\,2\,\mathrm{J}\dot{=} 138\,629\,\mathrm{J}\dot{=} 140\,\mathrm{kJ}\] \[p_2=\frac{p_1}{2}=\frac{2\cdot{ 10^5}}{2}\,\mathrm{Pa}= 10^5\,\mathrm{Pa}=100\,\mathrm{kPa}\] \[Q=W\dot{=}140\,\mathrm{kJ}\]#### Answer

During the isothermal expansion, the air has performed work of approximately 140 kJ. Heat supplied to the gas is 140 kJ as well, and the resultant pressure of the gas is 100 kPa.