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 V1 and final volume V2, 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 kg nitrogen mass T = 300 K nitrogen temperature p1 = 100 kPa = 1·105 Pa initial pressure of nitrogen p2 = 500 kPa = 5·105 Pa final pressure of nitrogen Q = ? given-off heat Table values:
Mm = 28 g mol−1 molar mass of nitrogen N2 R = 8.31 JK−1mol−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 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 that
\[W=\int\limits_{V_1}^{V_2}p\, \text{d}V,\]where V1 and V2 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 as a function of volume V, we use Boyle's Law:
\[p_1V_1=pV.\]From here we can express pressure p:
\[p = \frac{p_1V_1}{V}.\]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 V1 and final volume V2 from the equation of state for ideal gas
\[p_1V_1=\frac{m}{M_m}RT \qquad \Rightarrow \qquad V_1=\frac{mRT}{p_1M_m},\] \[p_2V_2=\frac{m}{M_m}RT \qquad \Rightarrow \qquad V_2=\frac{mRT}{p_2M_m}.\]After 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 is then determined as
\[Q= -W =-\frac{mRT}{M_m}\,\ln{\frac{p_1}{p_2}} =\frac{mRT}{M_m}\,\ln{\frac{p_2}{p_1}}.\]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 Q is equal to the absolute value of work performed by the nitrogen gas during compression. For this work it holds
\[W=\int\limits_{V_1}^{V_2}p\, \text{d}V,\]where V1 and V2 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 dT = 0 and we obtain
\[p\,\text{d}V=-V\text{d}p. \]If 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 p1 and p2 are initial and final pressures.
Now we need to express volume V as a function of pressure p. We use Boyle's Law
\[p_1V_1=pV, \]and from here we determine volume V:
\[V=\frac{p_1V_1}{p}.\]Unknown initial volume V1 is determined from the equation of state for ideal gas
\[p_1V_1=\frac{m}{M_m}RT. \]Therefore
\[V_1=\frac{mRT}{p_1M_m}.\]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 is then determined as follows
\[Q=-W=\frac{mRT}{M_m}\,\ln{\frac{p_2}{p_1}},\]which is the same relationship we obtained in the Solution section.
