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In an aquarium of a spherical shape with the radius of $r=10\;\mathrm{cm}$ which is completely filled with water, swim two identical fish in opposite directions. The fish has a cross-sectional area of $S=5\;\mathrm{cm}$, Newton's drag coefficient $C=0.2$ and it swims with a speed of $v=5\;\mathrm{km}\cdot h^{-1}$ relative to the water. How long have the fish to swim in the aquarium to increase the temperature of the water by 1 centigrade?

• All states of ideal gas can be shown on various diagrams: $pV$ diagram, $pT$ diagram and so on. The first quantity is shown is on vertical axis, the second on horizontal. Every point therefore determines 2 parameters. Sketch in a $pV$ diagram the 4 processes with ideal gas that you know. Do the same on a $Tp$ diagram. How would $UT$ diagram look like? Explain how would the unsuitability of these two variables appear on the diagram.

• What are the dimensions of entropy? What other quantities with the same dimensions do you know?

• In the text for this series we analysed a case of entropy increasing as heat flows into a gas. Perform a similar analysis for the case of heat flowing out of the gas.

• We know that entropy does not change during an adiabatic process. Therefore, the expression for entropy as a function of volume and pressure $S(p,V)$ can only contain a combination of pressure and volume that does not change during an adiabatic process.

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?

• Express the entropy of an ideal gas as functions $S(p,V)$, $S(T,V)$ and $S(U,V)$.

• Which types of processes (isobaric, isochoric, isothermal and adiabatic) can be reversible?
• Take the relation

$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?

• d gymnastics:
• Show that

$$\;\mathrm{d} (C f(x)) = C \mathrm{d} f(x)\,,$$

where $C$ is constant.

• Calculate

$$\;\mathrm{d} (x^2) \ \quad \mathrm{a} \quad \mathrm{d} (x^3).$$

• Show that

$$\;\mathrm{d}\left( \frac 1x \right)= -\frac {\mathrm{d} x}{x^2}$$

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$\$,.

• *Bonus: $This$ holds $$\sin \;\mathrm{d} \vartheta = \mathrm{d} \vartheta \quad a \quad \cos \mathrm{d} \vartheta = 1.$$ And you have the addition formula as well $$\sin (\alpha \beta ) = \sin \alpha \cos \beta \cos \alpha \sin \beta,$$ Prove $$\;\mathrm{d}\left( \sin \vartheta \right)=\, \mathrm{d} \vartheta \cos \vartheta .$$ * Bonus:** Similarly show

$$\;\mathrm{d} \left(\ln x \right)= \frac{\mathrm{d}x}{x}$$

using $$\ln (1 \;\mathrm{d} x) = \mathrm{d} x$$

• Explain, why isobaric temperature is lower than isochoric.

• As a little warm-up, to help you understand the numbers we'll be using, try to find height to what should be an average person (70 kg), lifted in order to use up all the energy of a standard Mars bar ( 250 cal for 50 g bar). Determine also what is the energy equivalent to $k_{B}T$ at room temperature and express it in electronvolts (i.e. the unit of energy equivalent to the kinetic energy electron gains when accelerated at potential difference of 1 V. Explicitly 1 eV = 1,602 \cdot 10^{-19} J).
• The Ideal Gas Law can be modified in many ways. If you rewrite it using amount of substance, instead of number of particles, you get $$pV = n N_\;\mathrm{A} k_\mathrm{B} T\,,$$ where $N_{A}k_{B}$ together is labeled as $R$ and is called universal gas constant. Express its value. Then modify the equation once again using mass of the gas and third time into a form containing gas density.
• Evaluate the volume of a single mole of gas at room temperature. It is useful to remember this number.
• And finally, a small consideration. Notice, when we were discussing the work of ideal gas, we automatically reached for the inner gas pressure value. Try to reason this choice of pressure. We might be objecting we should use the surrounding pressure or even the pressure difference between the inner and outer pressure. $Evaluation$ of this section will be moderate, do not be afraid to write whatever you think of yourself..</a>

Determine the dependency of the temperature of the solidification of the aqueous solution of sucrose at atmospheric pressure.

Into a container of volume $V=1\;\mathrm{m}$ in which there is a very low pressure (basically a prefect vacuum) we place $V_{0}=1l$ of water at room temperature $t_{0}$. What will be the final state in which the container and the water in it shall find itself in? For the purposes of the calculation assume that the container is prefectly thermally isolated from its surroundings and and has a negligible heat capacity.

Consider a tyre of a cylindrical shape and of a radius $R$ s an inner radius $rwidthd$ filled up to a pressure of $p$. We push down on the tyre with a force $F$. With this encumbrance the shape of the tyre changes from a cylinder to a cylindrical segment with the same inner and outer radius. Assume that the temperature of the tyre will not change. Determine the contact area of the tyre and road.

Consider a cylindrical container which fills the volume of $V=1l$. The container is closed with an airtight moving piston which has a non-negible mass $M$. Furthermore we know that the container is divided by horizontal partitions into $n$ sections and in the $i-thsection$ (it is numbered from the top ascendingly) there are 2^{$i}a$ particles, where $ais$ an undefined constant.The partitions are not fixed with regards to the container but at the same time they prevent the sections in which the ideal gas can be found from exchanging heat or particles. The whole system is at equilibrium. Then we make the mass of the piston twice as large and wait for equilibrium to arise again. How will the volume of the gas in the container change? Do not consider atmospheric pressure.

Test tubes of volumes 3 ml and 5 ml are connected by a short thin tube in which we can find a porous thermally non-conductive barrier that allows an equilbirum in pressures to be achieved within the system. Both test tubes in the beginning are filled with oxygen at a pressure of 101,25 kPa and a temperature of 20 ° C. We submerge the first test tube (3 ml) into a container which has a system of water and ice in equilbrium inside it and the other one (5 ml) into a container with steam. What wil the pressure be in the system of the teo test tubes be after achieving mechanical equilibrium? What would the pressure be if it would have been nitrogen and not oxygen that was in the test tubes?(while keeping other conditions the same)/p>

Take an empty cylindrical cup. Turn it upside down and push it beneath a calm water surface. How high will the column of air in the cup be depending on the submersion of the cup?