Lecture 16 - Temperature and Kinetic Theory

The SI unit of temperature is $K$ (Kelvin). We can convert a temperature from Celsius to Kelvin by adding $273$ (not exact, see below):

$$
T_K = T_C + 273
$$

It follows that $0^\circ C = 273K$ and $-273^\circ C = 0K$. The temperature of $0K$ is called
Absolute temperature could be identified by noticing how (at constant pressure) the volume of a gas decreases linearly with decreasing temperature. Extrapolating the linear relation shows that at $-273.15^\circ C$ the volume would drop to zero. Clearly the volume of a gas cannot be negative, which implies temperature cannot fall below $-273.15^\circ C$.

Using different types of gases, different amount, and running the experiment at different pressure all lead to the same temperature of zero volume, so clearly there is something special about $-273.15^\circ C$, which we now call absolute zero. The Kelvin scale is defined so that absolute zero is exactly $0K$.

One can see in the Kelvin scale, $V\propto T$.

One can also plot the variation of gas pressure with temperature in a container at constant volume. Once again, the lines all crosses the $T$-axis at absolute zero. Therefore we find that $P\propto T$ in the Kelvin scale.

An ideal gas is a gas of point particles which has no other interactions besides collision among molecules. There is no intermolecular force. An ideal gas obeys the ideal gas law:

$$
PV = nRT
$$

where $R=8.31JK^{-1}mol^{-1}$ is the ideal gas constant, $n$ is the number of moles.
$P$ | $Pa$ |

$V$ | $m^3$ |

$T$ | $K$ |

$n$ | $mol$ |

Real gas molecules have non-zero volume (i.e. not point particles), and there is generally intermolecular force. However, for most gases at room temperature the size of the gas molecules are small compared to the whole volume of the gas, and the molecules are usually separated far enough for the intermolecular forces to be negligible (except when the gas is close to phase transition) so the ideal gas law usually works very well.

$n$ and $N$ are related to each other via $N= n N_A$, where $N_A = 6\times 10^{23}$ is the Avogadro's number. $N_A$ is the number of molecules in 1 mole of gas.

A good analogy to Avogadro's number is the concept of *dozen* used in a bakery. You can think of $N_A^{\text{baker}} = 12$ for a baker. So if you buy 3 dozens of donuts, it means you are buying $N = nN_A = (3)(12) = 36$ donuts.

Number of moles, $n$ | Number of molecules, $N = n N_A$ |
---|---|

$1$ | $6\times 10^{23}$ |

$2$ | $12\times 10^{23}$ |

$3$ | $18\times 10^{23}$ |

$4$ | $24\times 10^{23}$ |

Number of dozens, $n$ | Number of donuts, $N = n N_A$ |
---|---|

$1$ | $12$ |

$2$ | $24$ |

$3$ | $36$ |

$4$ | $48$ |

The SI unit for volume is $m^3$. Unit conversion is a very basic skill for a science major. Make sure you know how to convert units to and from $cm^3$ and $L$ (liter) correctly or you will be penalized heavily. Here is a quick reminder: $$ \begin{eqnarray} 1m &=& 100cm \tag{length}\\ \Rightarrow 1m^2 &=& (100cm)^2 = 10^4 cm^2 \tag{area}\\ \Rightarrow 1m^3 &=& (100cm)^3 = 10^6 cm^3 \tag{volume}\\ \end{eqnarray} $$ Using the above, we get: $$ \begin{eqnarray} 1cm^3 &=& 10^{-6} m^3 \\ 1L &=& 1000cm^3 = 1000\times 10^{-6}m^3 = 10^{-3} m^3 \end{eqnarray} $$

The SI unit of pressure is $Pa$ (Pascals), which is equivaldent to $N/m^2$.

A common non-SI unit for pressure is $atm$, or "atmospheric pressure". To convert to SI unit, use $1atm \approx 10^5Pa$.

The term **STP** stands for **standard temperature and pressure**, meaning $T=0^\circ C$ and $P=1atm$.

An alternative way of writing the ideal gas law is $PV = NkT$, where $k=1.38\times 10^{-23}J/K$ is the Boltzmann constant. Compare with the version presented above $PV = nRT$, we could see: $$ \begin{eqnarray} NkT &=& nRT \\ \Rightarrow Nk &=& nR \\ \Rightarrow k &=& \frac{n R}{N} = \frac{n R}{n N_A} \\ &=& \frac{R}{N_A} \\ &=& 1.38 \times 10^{-23} J/K \end{eqnarray} $$

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A useful equation derived from the ideal gas law is:

$$
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
$$

The implicit assumption here is $n_1 = n_2$. In other words, the container of the box is not leaking so the amount of gas stays constant. For the proof, start with $PV = nRT \Rightarrow nR = \frac{PV}{T}$. The assumption $n_1 = n_2$ then gives the equation above.
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Name | Symbol | Unit | Meaning |
---|---|---|---|

Pressure | $P$ | $Pa$ | force per unit area |

Volume | $V$ | $m^3$ | size of a gas |

Temperature | $T$ | $K$ | how hot an object is |

Avogadro's number | $N_A = 6\times 10^{23}$ | none | number of molecules in one mole |

Number of moles | $n$ | $mol$ | amount of molecules as multiples of $N_A$ |

Number of molecules | $N$ | none | $N = nN_A$ |