$\frac{dP}{dy}=-\rho g$

$\int_{P_{1}}^{P_{2}}\,dP=-\int_{y_{1}}^{y_{2}}\rho g\,dy$

$P_{2}-P_{1}=-\rho g(y_{2}-y_{1})$

The depth $h=y_{2}-y_{1}$ and the pressure at the top $P_{2}$ is atmospheric pressure $P_{0}$

We can therefore say that the pressure at depth $h$, $P=P_{1}$ is

$P=P_{0}+\rho gh$

We can see that we need to take in to account the pressure of the atmosphere above the container!

If you measure the pressure in your tires you are actually measuring the pressure difference between the atmospheric pressure. The absolute pressure is therefore the sum of the gauge pressure $P_{G}$, which is what you measure, and the atmospheric pressure $P_{0}$

$P=P_{0}+P_{G}$

In a fluid of uniform density $\rho_{f}$

$F_{B}=F_{2}-F_{a}=\rho_{f}gA(h_{2}-h_{1})=\rho_{f}gA\Delta h=\rho_{f}Vg=m_{f}g$

We can note from this equation that the buoyant force does not depend on the depth of the object or on the density of the object. The buoyant force is determined by the weight of the fluid which is displaced by the object.

$m\frac{d^{2}x}{dt^{2}}=-kx$

$x=A\cos(\omega t+\phi)$

$A$ is the amplitude, $\omega=\frac{2\pi}{T}=2\pi f$ and $\phi$ allows us to change the starting point of the motion.

$\frac{dx}{dt}=v=-\omega A\sin(\omega t+\phi)$

$\frac{d^{2}x}{dt^{2}}=a=-\omega^{2}A\cos(\omega t+\phi)$

$-m\omega^{2}A\cos(\omega t+\phi)=-kA\cos(\omega t+\phi)$

which is true if

$\omega^{2}=\frac{k}{m}$

$T=2\pi\sqrt{\frac{m}{k}}$

If we freeze a wave at a certain time the displacement of the points can often be represented by a sinusoidal function, $D(x)=A\sin\frac{2\pi}{\lambda}x$.

If the wave moves so that it takes a time $T$, the period for a wavelength $\lambda$ to pass a point we can say that the velocity of a wave is $v=\frac{\lambda}{T}=f\lambda$.

In order to have a wave which at time $t=0$ has a displacement function $D(x)=\sin\frac{2\pi}{\lambda}x$ propagate with $v$ we can write

$D(x,t)=A\sin\frac{2\pi}{\lambda}(x-vt)$

which using $v=\frac{\lambda}{T}=f\lambda$ can be written

$D(x,t)=A\sin(\frac{2\pi}{\lambda}x-\frac{2\pi t}{T})=A\sin(kx-\omega t)$

$k$ is the wave number, $k=\frac{2\pi}{\lambda}$ and the angular frequency $\omega=2\pi f$

The velocity of the waves propagation, which we call the phase velocity can be expressed in terms of $\omega$ and $k$

$v=f\lambda=\frac{\omega}{2\pi}\frac{2\pi}{k}=\frac{\omega}{k}$

The intensity is the average power per unit area

$I=\frac{\bar{P}}{S}$

As a 3 dimensional wave propagates from a point the area through which the wave passes increases, when the power out put is constant, the intensity decreases as $\frac{1}{r^{2}}$

$\frac{I_{2}}{I_{1}}=\frac{\bar{P}/4\pi r_{2}^{2}}{\bar{P}/4\pi r_{1}^{2}}=\frac{r_{1}^2}{r_{2}^{2}}$

In order for the power output to be constant the amplitude must also decrease as $S_{1}A_{1}^2=S_{2}A_{2}^2$ implying $A\propto\frac{1}{r}$

Our sensitivity to the loudness of sound is logarithmic, a sound that is ten time as intense sounds only twice as loud to us. The sound level $\beta$ is thus measured on a logarithmic scale in decibels is

$\beta=10\log_{10}\frac{I}{I_{0}}$

$I_{0}$ is the weakest sound intensity we can hear $I_{0}=1.0\times 10^{-12}\mathrm{W/m^{2}}$

Beats occur when two waves with frequencies close to one another interfere.

If the two waves are described by

$D_{1}=A\sin2\pi f_{1}t$

and

$D_{2}=A\sin2\pi f_{2}t$

$D=D_{1}+D_{2}$

$D=2A\cos2\pi(\frac{f_{1}-f_{2}}{2})t\sin2\pi(\frac{f_{1}+f_{2}}{2})t$

A general formula for the doppler effect is

$f'=f\frac{(v_{sound}\pm v_{obs})}{(v_{sound}\mp v_{source})}$

Top part of the $\pm$ or $\mp$ sign is for an source or observer moving towards each other, the bottom part is for motion away from each other.

If two object with different temperatures are brought in to contact with one another thermal energy will flow from one to another until the temperatures are the same, and we then say that the objects are in thermal equilibrium.

The zeroth law of thermodynamics states that:

“If two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.”

Most, but not all, materials expand when heated. The change in length of material due to linear thermal expansion is

$\Delta l=\alpha l_{0}\Delta T$

$\alpha$ is the coefficient of linear expansion of the material, measured in $\mathrm{(^{o}C)^{-1}}$

The length of the object after it's temperature has been changed by $\Delta T$ is

$l=l_{0}(1+\alpha\Delta T)$

A material expands in all directions, and if we are interested in the volume changes of a rectangular object, that is isotropic, meaning it expands in the same way in all directions, then

$\Delta V = \beta V_{0}\Delta T$

$V_{0}=l_{0}w_{0}h_{0}$ → $V=l_{0}(1+\alpha\Delta T)w_{0}(1+\alpha\Delta T)h_{0}(1+\alpha\Delta T)$

$\Delta V=V-V_{0}=V_{0}(1+\alpha\Delta T)^{3}-V_{0}=V_{0}[3(\alpha\Delta T)+3(\alpha\Delta T)^{2}+(\alpha\Delta T)^{3}]$

If $\alpha\Delta T << 1$ then $\beta \approx 3\alpha$

A mole of gas is a given number of molecules, Avagadro's number, $N_{A}=6.02\times 10^{23}$. If we have a certain mass $m$ of a gas which has a certain molecular mass (measured in atomic mass units, $\mathrm{u}$, which are also the number of grams per mole.), the the number of moles $n$ is given by

$n=\frac{m[\mathrm{g}]}{\textrm{molecular mass}[\mathrm{g/mol}]}$

and

$PV=nRT$ where $R=8.314\mathrm{J/(mol.K)}$

The ideal gas law can also be written in terms of the number of molecules

$PV=nRT=\frac{N}{N_{A}}RT=NkT$

where $k=\frac{R}{N_{A}}=\frac{8.314\mathrm{J/(mol.K)}}{6.02\times 10^{23}}=1.38\times 10^{-23}\mathrm{J/K}$ is the Boltzmann Constant.

The average kinetic energy of molecules in an monatomic ideal gas is

$\bar{K}=\frac{1}{2}m\bar{v^{2}}=\frac{3}{2}kT$

$f(v)=4\pi (\frac{m}{2\pi k T})^{\frac{3}{2}}v^{2}e^{-\frac{1}{2}\frac{mv^{2}}{kT}}$

A quantity of heat, $Q$, flowing into an object leads to a change in the temperature of the object, $\Delta T$, which is proportional to it's mass $m$ and a characteristic quantity of the material, it's specific heat, $c$

$Q=mc\Delta T$

We can see that heat flowing in to an object is positive $\Delta T>0$ and heat flowing out is negative $\Delta T < 0$

The specific heat is the heat capacity per a unit of mass, in SI the units of specific heat are $\mathrm{\frac{J}{kg.K}}$.

The assumption of an isolated system is very useful in problem solving as it says that the sum of the heat transfers in the system must be zero.

$\Sigma Q = 0$

In a system where the different objects start at different temperatures, but eventually come to an equilibrium temperature $T$

$\Sigma Q = m_{1}c_{1}(T-T_{i1})+m_{2}c_{2}(T-T_{i2})+..$

Phase changes from a low temperature phases to a high temperature phase require a certain amount of heat, called the latent heat.

The latent heat of of fusion, $L_{f}$, refers to a change from solid to liquid and the latent heat of vaporization, $L_{v}$, refers to a change from liquid to gas. The heat required to change a mass $m$ of a substance from one phase to another is

$Q=mL$

During a change from one phase to another the temperature of the system remains constant.

The first law of thermodynamics, dictates how internal energy, heat and work are related to each other. For a closed system the first law states that the change in the internal energy of a system, $\Delta E_{int}$, is the sum of the heat added **to** the system $Q$ and the net work done **by** the system $W$.

$\Delta E_{int}=Q-W$

The table shows some of the results that apply to a particular kind of thermal process

Process | Constant | ΔE_{int} | Q | W |
---|---|---|---|---|

Isothermal | T | 0 | Q=W | W=Q |

Isobaric | P | Q-PΔV | ΔE_{int}+PΔV | PΔV |

Isovolumetric | V | Q | ΔE_{int} | 0 |

Adiabatic | -W | 0 | -ΔE_{int} |

For an ideal gas $P=\frac{nRT}{V}$, so for an isothermal process

$W=\int_{V_{A}}^{V_{B}}P\,dV=nRT\int_{V_{A}}^{V_{B}}\frac{dV}{V}=nRT\ln\frac{V_{B}}{V_{A}}$

In an isobaric process the pressure is constant so the work is

$W=\int_{V_{A}}^{V_{B}}P\,dV=P(V_{B}-V_{A})=P\Delta V$

and if the system is an ideal gas

$W=P(V_{B}-V_{A})=nRT_{B}(1-\frac{V_{A}}{V_{B}})=nRT_{A}(\frac{V_{B}}{V_{A}}-1)$

In an isovolumetric process the work done is zero

$W=0$

For an ideal gas $c_{P,m}-c_{V,m}=R$

The equipartition theorem can be used to explain the higher heat capacity of more complicated gases. The equipartition theorem states that energy is equally shared between the different degrees of freedom the molecules in the gas have.

Each degree of translational or rotational freedom contributes $\frac{1}{2}R$ to the molar specific heat at constant volume $c_{V,m}$

$PV^{\gamma}=\mathrm{constant}$

$\gamma=\frac{c_{P,m}}{c_{V,m}}$

**Heat cannot spontaneously flow from a cold object to a hot one, whereas the reverse, a spontaneous flow of heat from a hot object to a cold one, is possible.**

or

**No device is possible whose sole effect is to transform a given amount of heat directly in to work.**

$Q_{H}=W+Q_{L}$

The efficiency, $e$, of an engine is defined as the ratio of the work we get from the engine $W$ to the input heat $Q_{H}$

$e=\frac{W}{Q_{H}}$

$e=\frac{W}{Q_{H}}=\frac{Q_{H}-Q_{L}}{Q_{H}}=1-\frac{Q_{L}}{Q_{H}}$

The coefficient of performance, $COP$, of a refrigerator is defined as the heat removed $Q_{L}$ divided by the work done $W$. As before we apply the first law $Q_{L}+W=Q_{H}$ so

$COP=\frac{Q_{L}}{W}=\frac{Q_{L}}{Q_{H}-Q_{L}}$

As with a heat engine we can consider the Carnot cycle to be the ideal case, which means that for an ideal refrigerator

$COP=\frac{T_{L}}{T_{H}-T_{L}}$

We will define the change in entropy in a reversible process at constant temperature as

$\Delta S =\frac{Q}{T}$

If we want to treat non-constant temperature cases we can express the change of entropy in differential form

$dS=\frac{dQ}{T}$

and then the change in entropy in going from state $a$ to $b$ will be

$\Delta S =S_{b}-S_{a}=\int_{a}^{b}\,dS=\int_{a}^{b}\frac{dQ}{T}$

A state variable is a variable that describes the current state of a dynamical system. Heat is not a state variable, it depends on the path taken to get the system to it's state. Entropy, on the other hand, is a state variable, the change in entropy required to change a system from one state to another via a reversible process is independent of the path taken. If we want to find the entropy change for an irreversible process (ie. any real process) that goes from state A to state B we can calculate the entropy change for a reversible process that goes from point to A and B, and the entropy change for the system will be the same for the two processes. The difference between the reversible and irreversible process is that in a reversible process there is no increase in the entropy of the environment, but in an irreversible process there is.