Boyle's Law | Combined Gas Law | |

Charles' Law | Ideal Gas Law | |

Gay-Lussac's Law | Dalton's Law | |

Avogadro's Law | Graham's Law | |

Diver's Law | Return to KMT & Gas Laws Menu |

As far as the ChemTeam knows, this law does not have a name because it was never really "discovered."

Gives the relationship between amount and temperature when pressure and volume are held constant. Remember, amount is measured in moles. Also, since volume is one of the constants, that means the container holding the gas is rigid-walled and inflexible.

If the amount of gas in a container is increased, the temperature must decrease.

If the amount of gas in a container is decreased, the temperature must increase.

Why?

Suppose the amount is increased. This means there are more gas molecules and this will increase the number of impacts on the container walls. This means the gas pressure inside the container will increase, since the volume is constant. However, this cannot be allowed since pressure also remains constant. Lowering the temperature will cause the molecules to move slower, thus lessening the impacts on the container walls and reducing the pressure to the original value.

The mathematical form of the "no-name" Law is:

nT = k

Let n_{1} and T_{1} be a amount-temperature pair of data at the start of an experiment. If the amount is changed to a new value called n_{2}, then the temperature will change to T_{2}.

We know this:

n_{1}T_{1}= k

And we know this:

n_{2}T_{2}= k

Since k = k, we can conclude the following:

n_{1}T_{1}= n_{2}T_{2}.

This equation of n_{1}T_{1} = n_{2}T_{2} will be very helpful in solving "no-name" Law problems. If you ever see one!

The "no-name" Law is an inverse mathematical relationship. This means that, as one variable (either n or T) changes in value, the other changes in the opposite direction. The constant k remains the same value.

In March 2017, I finally saw a "no-name" gas law problem. A teacher in New Jersey was kind enough to send me one he has used. Here it is:

**Example #1:** A hot air balloon is filled with 118,000 moles of air at 20.0 °C. If the air in this balloon is heated to 98.0 °C, how many moles of air will now be in the hot air balloon? (The volume and pressure of the air in the balloon stays constant.)

**Solution:**

1) Write the "no-name" law:

n_{1}T_{1}= n_{2}T_{2}

2) Insert values:

(1.18 x 10^{5}mol) (293 K) = (n_{2}) (371 K)

3) Solve:

n_{2}= 93191.4 moles of airRounded to three sig figs, the answer is 93200 mole (or 9.32 x 10

^{4}mol)

My thanks to L.

Boyle's Law | Combined Gas Law | |

Charles' Law | Ideal Gas Law | |

Gay-Lussac's Law | Dalton's Law | |

Avogadro's Law | Graham's Law | |

Diver's Law | Return to KMT & Gas Laws Menu |