You can find the serial also in the yearbook.
We are sorry, this serial has not been translated.
$T=\frac{pV}{nR}\$,,
where $n=1mol$, $p=100kPa$ and $V=22l$. How will $T$ change, if we change both $p$ and $V$ by 10$%$, by 1$%$ or by 0$,1%?$ Calculate it in two ways: precisely and by using the relation: $$\;\mathrm{d} T=T_{,p} \mathrm{d} p T_{,V} \mathrm{d} V .$$
What is the difference between the results?
$$\;\mathrm{d} (C f(x)) = C \mathrm{d} f(x)\,,$$
where $C$ is constant.
$$\;\mathrm{d} (x^2) \ \quad \mathrm{a} \quad \mathrm{d} (x^3).$$
$$\;\mathrm{d}\left( \frac 1x \right)= -\frac {\mathrm{d} x}{x^2}$$
from the definition, that is $$\;\mathrm{d} \left(\frac 1x \right)= \frac {1}{x \mathrm{d} x} - \frac 1x$$
This might be handy: $(x \;\mathrm{d} x)(x-\mathrm{d}$ x) = x^2 - (\mathrm{d} x)^2 = x^2$\$,.
$$\;\mathrm{d} \left(\ln x \right)= \frac{\mathrm{d}x}{x}$$
using $$\ln (1 \;\mathrm{d} x) = \mathrm{d} x$$
What is this expression? Draw lines of constant entropy on a $pV$ diagram ($p$ on vertical axis, $V$ on horizontal). Does this agree with the expression for entropy we have derived?
$$\Delta S_{tot} \ge 0 }$$
and given the equation from the text of the serial
$$\Delta S_{tot} = \frac{-Q}{T_H} \frac{Q-W}{T_C}$$
express $W$ and derive this way the inequality for work
$$W\le Q\left( 1 - \frac {T_C}{T_H} \right).$$
Hint: Write out 4 equations connecting 4 vertices of the Carnot cycle
$$p_1 V_1 = p_2 V_2 $$
$$p_2 V_2^{\kappa} = p_3V_3^{\kappa}$$
$$p_3V_3 = p_4V_4$ p_4V_4^{\kappa} = p_1V_1^{\kappa}$
and multiply all of them together. By modifying this equation you should be able to get
$$\frac {V_2}{V_1} = \frac {V_3}{V_4}.$$
Next step is using the equation for the work done in an isothermal process: when going from the volume $V_{A}$ to the volume $V_{B}$, the work done on a gas is
$nRT\,\;\mathrm{ln}\left(\frac{V_A}{V_B}\right)$.
Now the last thing we need to realize is that the work in an isothermal process is equal to the heat (with the correct sign) a calculate the work done by the gas (there is no contribution from the adiabatic processes) and the heat taken away.
$ For the correct solution, you only need to fill in the details.$
$p=p_0\;\mathrm{e}^{-\frac{V}{V_0}}$,
where $p_{0}$ and $V_{0}$ are constants. Show for which values of $V$ (during the expansion) the heat is going into the gas and for which out of it.
$$S(U, V, N) = \frac{s}{2}n R \ln \left( \frac{U V^{{\kappa} -1}}{\frac{s}{2}R n^{\kappa} } \right) nR s_0$$
and the relation for the change of the entropy
$$\;\mathrm{d} S = \frac{1}{T}\mathrm{d} U \frac{p}{T} \mathrm{d} V - \frac{\mu}{T} \mathrm{d} N$$
to calculate chemical potential as a function of $U$, $VaN$. Modify it further to get the function of $T$, $pandN$.
Hint: The coefficients like 1 ⁄ $T$ in front of d$U$ can be calculated as a partial derivative of $S(U,V,N)$ by $U$. Don't forget that ln$(a⁄b)=\lna-\lnb$ and that $n=N⁄N_{A}$.
Bonus: Express similarly temperature and pressure as functions of $U$, $VandN$. Eliminate the pressure dependence to get the equation of state.
Hint: To calculate the work, this equation can be useful:
$$\int _{a}^{b} \frac{1}{x} \;\mathrm{d}x = \ln \frac{b}{a}.$$
$$2\,\;\mathrm{H}_2 \mathrm{O}_2\longrightarrow2\,\mathrm{H}_2\mathrm{O},$$
where both the reactants and the product are gases at standard conditions. Find the change of entropy in this reaction. Give results per mole.
$j=\frac{3}{4}\frac{k_\;\mathrm{B}^4\pi^2}{45\hbar^3c^3}cT^4$.
Substitute the values of the constants and compare the result with the Stefan-Boltzmann law.
Hint: The law for an adiabatic process with an ideal gas was derived in the second part of this series (Czech only).
$$\delta Q / T = f_{,T} \;\mathrm{d} T f_{,V} \mathrm{d} V\,,$$
then functions $f_{,T}$ and $f_{,V}$ obey the necessary condition for the existence of entropy, that is
$$\frac{\partial f_{,T}(T, V)}{\partial V} = \frac{\partial f_{,V}(T, V)}{\partial T} $$