## Heat Given Off by Nitrogen

### Task number: 3950

Mow much heat is it necessary to be given off by 56 g of nitrogen in order to isothermally compress it from the pressure of 100 kPa to the pressure of 500 kPa at the temperature of 300 K?

Note: Consider nitrogen to be an ideal gas.

#### Hint 1

Realize that at constant temperature, the internal energy of ideal gas does not change and the given-off heat is therefore equal to work performed by the surroundings during the compression of nitrogen gas, or rather the absolute value of work performed by nitrogen (during compression the performed work is negative).

#### Hint 2

We need to determine the performed work using integral calculus due to the fact that the pressure is not constant.

#### Hint 3

To determine pressure

*p*as a function of volume*V*(and conversely), we can use Boyle's Law for isothermal process.#### Hint 4

To determine initial volume

*V*_{1}and final volume*V*_{2}, use the equation of state for ideal gas.#### Analysis

Since at constant temperature the internal energy of the gas does not change, from the First Law of Thermodynamic it follows that the given-off heat is equal to the absolute value of work performed by the nitogen gas during its compression. To determine it we need to use integral calculus due to the fact that pressure and volume are not constant.

The pressure of nitrogen gas as a volume function is determined from Boyle's Law that applies for isothermal processes and perform an itegration of resulting function with respect to volume.

After expressing the unknown initial volume and final volume we finally use the equation of state for ideal gas.

#### Given Values

*m*= 56 g = 0.056 kgnitrogen mass *T*= 300 Knitrogen temperature *p*_{1}= 100 kPa = 1·10^{5}Painitial pressure of nitrogen *p*_{2}= 500 kPa = 5·10^{5}Pafinal pressure of nitrogen *Q*= ?given-off heat *Table values:**M*_{m}= 28 g mol^{−1}molar mass of nitrogen N _{2}*R*= 8.31 JK^{−1}mol^{−1}molar gas constant #### Solution

During isothermal process the internal energy of the gas does not change which according to the First Law of Thermodynamics means that the given-off heat

\[W=\int\limits_{V_1}^{V_2}p\, \text{d}V,\]*Q*is equal to work*W*' performed by the surroundings during gas compression, or rather the absolute value of work*W*performed by the gas (it is negative during compression). It is true thatwhere

*V*_{1}and*V*_{2}are initial and final volume of the gas and*p*is the gas pressure that continuously changes during the compression (it is a volume function).To determine pressure

\[p_1V_1=pV.\]*p*as a function of volume*V*, we use Boyle's Law:From here we can express pressure

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

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

\[=p_1V_1 \int\limits_{V_1}^{V_2}\frac{1}{V}\, \text{d}V = \]we perform the integration and substitute the limits

\[=p_1V_1[\ln V]_{V_1}^{V_2} = p_1V_1 \ln \frac{V_2}{V_1}.\]Now we determine the unknown initial volume

\[p_1V_1=\frac{m}{M_m}RT \qquad \Rightarrow \qquad V_1=\frac{mRT}{p_1M_m},\] \[p_2V_2=\frac{m}{M_m}RT \qquad \Rightarrow \qquad V_2=\frac{mRT}{p_2M_m}.\]*V*_{1}and final volume*V*_{2}from the equation of state for ideal gasAfter substitution we obtain

\[W=p_1\frac{mRT}{p_1M_m}\,\ln{\frac{\frac{mRT}{p_2M_m}}{\frac{mRT}{p_1M_m}}}=\frac{mRT}{M_m}\,\ln{\frac{p_1}{p_2}}.\]The given-off heat

\[Q= -W =-\frac{mRT}{M_m}\,\ln{\frac{p_1}{p_2}} =\frac{mRT}{M_m}\,\ln{\frac{p_2}{p_1}}.\]*Q*is then determined as#### Numerical Solution

\[Q=\frac{mRT}{M_m}\,\ln{\frac{p_2}{p_1}}=\frac{0.056\cdot{8.31}\cdot{300}}{0.028}\cdot \ln{\frac{500}{100}}\,\mathrm{J}\dot{=}8025\,\mathrm{J}\dot{=}8\,\mathrm{kJ}\]#### Answer

It is necessary to give off heat of approximately 8 kJ.

#### Alternative Solution

As is stated in the solution of this task, the given-off heat

\[W=\int\limits_{V_1}^{V_2}p\, \text{d}V,\]*Q*is equal to the absolute value of work performed by the nitrogen gas during compression. For this work it holdswhere

*V*_{1}and*V*_{2}are initial and final volume of the gas and*p*is pressure of the gas.Given the fact that we know initial and final pressure of the gas and not its volumes, we adjust this formula in such a way that the work would be determined by volume as a pressure function.

We start with the equation of state

\[pV=nRT,\]that we differentiate:

\[p\,\text{d}V+V\text{d}p=nR\,\text{d}T.\]Since it is an isothermal process, we can say that d

\[p\,\text{d}V=-V\text{d}p. \]*T*= 0 and we obtainIf we substitute this expression to our original integral expression for work and adjust the limits of the integral, we obtain

\[W=-\int\limits_{p_1}^{p_2}V\, \text{d}p,\]where

*p*_{1}and*p*_{2}are initial and final pressures.Now we need to express volume

\[p_1V_1=pV, \]*V*as a function of pressure*p*. We use Boyle's Lawand from here we determine volume

\[V=\frac{p_1V_1}{p}.\]*V*:Unknown initial volume

\[p_1V_1=\frac{m}{M_m}RT. \]*V*_{1}is determined from the equation of state for ideal gasTherefore

\[V_1=\frac{mRT}{p_1M_m}.\]Now we can perform the integration:

\[W=-\int\limits_{p_1}^{p_2}V\, \text{d}p = -\int\limits_{p_1}^{p_2}\frac{p_1V_1}{p}\text{d}p = -\int\limits_{p_1}^{p_2}\frac{p_1mRT}{pp_1M_m}\text{d}p = \]we factor the constants out of the integral

\[=-\frac{mRT}{M_m}\int\limits_{p_1}^{p_2}\frac{1}{p}\text{d}p = \]we perform the integration and substitute the limits

\[=-\frac{mRT}{M_m}\,\left[\ln p\right]_{p_1}^{p_2}= -\frac{mRT}{M_m}\,\ln{\frac{p_2}{p_1}}.\]The given-off heat

\[Q=-W=\frac{mRT}{M_m}\,\ln{\frac{p_2}{p_1}},\]*Q*is then determined as followswhich is the same relationship we obtained in the Solution section.