## The Gas Equation

**The three gas laws give the following equations:**

**pV = constant**

*(when T is kept constant)*

V |
= constant (when p is kept constant) |

T |

p |
= constant (when V is kept constant) |

T |

**These 3 equations are combined to give the ideal gas equation:**

pV |
= constant |

T |

Where,

p = the pressure of the gas

V = the volume the gas occupies

T = the gas temperature on the Kelvin scale

From this equation we know that if a fix mass of gas has starting values of p1, V1 and T1, and then some time later has value p2, V2 and T2, the equation can be written as:

p_{1}V_{1} |
= |
p_{2}V_{2} |

T_{1} |
T_{2} |

## Example:

Sabah pumps up her front bicycle tyre to 1.7 x 10^{5} Pa. The volume of air in the tyre at this pressure is 300 cm^{3}. She takes her bike for a long ride during which the temperature of the air in the tyre increases from 20°C to 30°C. Calculate the new front tyre pressure assuming the tyre had no leaks and so the volume remained constant?

### Solution:

p

_{1}= 1.7 x 10^{5}Pa

T_{1}= 20°C = 20 + 273 = 293 K

V_{1}= 300 cm^{3}p

_{2}= ?

T_{2}= 30°C = 30 + 273 = 303 K

V_{2}= 300 cm^{3}

Using:

p_{1}V_{1}=p_{2}V_{2}T_{1}T_{2}

Rearranging for P2

p_{2}=p_{1}V_{1}T_{2}V_{2}T_{1}

We know **V1 = V2**, therefore the equation can be simplified to:

p_{2}=p_{1}T_{2}T_{1}

p_{2}=1.7^{}x 10^{5}x 303293 ^{}

p1.76 x 10_{2}=^{5}Pa