Answer to Question #121927 in Molecular Physics | Thermodynamics for christopher seebaran

Question

Ten moles of a monaatomic ideal gas initially has a volume of 0.800m3. The gas undergoes four processes: 1) without changing its temperature, the gas expands to a volume of 1.200m3, at which point its pressure is 20.00KPa, 2) The gas expands to a volume of 1.480m3 without heat entering or leaving the gas. 3) without changing its temperature, the gas is compressed until its pressure is 21.15KPa. 4) The gas is compressed to a pressure of 30.00KPa without heat entering or leaving the gas.
I) calculate the work done by the gas in processes 1) and 3).
II) Determine the process that the gas undergoes
III) Using the definition of the entropy, calculate the total change in entropy during the four processes. Comment on your answer.

Answer 1

If we consider the ideal monoatomic gas, we'll see that the adiabatic index $\gamma = \dfrac53.$

Before the processes $V_0 = 0.800\,\mathrm{m}^3, \nu=10\,\mathrm{mol}.$ $p_0V_0 =\nu R T_0.$

After the first process $p_1V_1=\nu RT_0$ , therefore $p_0 =\dfrac{p_1V_1}{V_0} = \dfrac{2\cdot10^4\,\mathrm{Pa}\cdot1.200\,\mathrm{m}^3}{0.800\,\mathrm{m}^3} = 3\cdot10^4\,\mathrm{Pa}.$ The temperature is $T_0 = \dfrac{p_1V_1}{\nu R} = \dfrac{3\cdot10^4\,\mathrm{Pa}\cdot0.800\,\mathrm{m}^3}{10\cdot8.31} \approx 289\,\mathrm{K}.$

The second process is adiabatic, so

$p_1V_1^{\gamma}=p_2V_2^{\gamma} \Rightarrow p_2 =p_1\left(\dfrac{V_1}{V_2} \right)^{\gamma} = 2\cdot10^4\,\mathrm{Pa}\cdot\left( \dfrac{1.200}{1.480}\right)^{5/3} = 14100\,\mathrm{Pa}.$

The third process is isothermal, $p_2V_2 = p_3V_3, \;\; V_3 = \dfrac{p_2V_2}{p_3} = \dfrac{14100\,\mathrm{Pa}\cdot1.480\,\mathrm{m}^3}{21.15\cdot10^3\,\mathrm{Pa} } = 0.987\,\mathrm{m^3}.$

The temperature is $T_2 = \dfrac{p_3V_3}{\nu R} = \dfrac{21150\cdot0.987\,\mathrm{m}^3}{10\cdot8.31} = 251\,\mathrm{K}.$

The last process is adiabatic, so $p_3V_3^{\gamma} = p_4V_4^{\gamma} \Rightarrow V_4 = V_3\left(\dfrac{p_3}{p_4} \right)^{1/\gamma} = 0.8\,\mathrm{m}^3.$

We can see that $V_4=V_0, \; p_4=p_0,$ so the $T_4=T_0.$ The process is cyclic, it is a kind of Carnot cycle (see https://en.wikipedia.org/wiki/Carnot_cycle).

i) 1) and 3) processes are isothermal, so the work is equal to the heat.

$A_1 = \int\limits_{V_0}^{V_1}p\,dV =\nu \int\limits_{V_0}^{V_1}\dfrac{RT_0}{V}\,dV =\nu RT_0\ln\dfrac{V_1}{V_0} = 10\cdot8.31\cdot 289\cdot\ln\dfrac{1.2}{0.8} = 9740\,\mathrm{J}.$

$A_3 =\int\limits_{V_2}^{V_3}p\,dV =\nu \int\limits_{V_2}^{V_3}\dfrac{RT_3}{V}\,dV =\nu RT_3\ln\dfrac{V_3}{V_2} = 10\cdot8.31\cdot 251\cdot\ln\dfrac{0.987}{1.48} = -8450\,\mathrm{J}.$

ii) The process is cyclic, it is a kind of Carnot cycle (see https://en.wikipedia.org/wiki/Carnot_cycle).

iii) The change of entropy in isothermal process is (see https://en.wikipedia.org/wiki/Isothermal_process)

$\Delta S_1 = \nu R \ln \dfrac{P_1}{P_0}=10\cdot8.31\cdot\ln \dfrac{2\cdot10^4}{3\cdot10^4} = -33.7\,\mathrm{J/K}.$

$\Delta S_3= \nu R \ln \dfrac{P_3}{P_2}=10\cdot8.31\cdot\ln \dfrac{21150}{14100} = 33.7\,\mathrm{J/K}.$

In 2) and 4) adiabatic processes $\Delta S_2=\Delta S_4 = 0.$

So the entropy is constant in this cyclic process.