For a typical geyser, underground water seeps into a deep narrow shaft in the ground and is heated from below. Because of the depth, the pressure on the water is high so the water at the bottom of the shaft boils at a much higher temperature than normal.

What happens to the volume of a 1.0 L bubble of water vapour at 130 degrees celcius and 305 kilopascals when it reaches the surface, where the conditions are 93 degrees celcius and 101 kilopascals?

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For this, we can use the traditional combined gas law.

`(p_1 V_1)/(T_1) = (p_2 V_2)/(T_2)`

Given:

- p1 = 305 kilopascals
- V1 = 1.0 liters
- T1 = 130 + 273.15 = 403.15 K
- p2 = 101 kilopascals
- T2 = 93 + 273.15 = 366.15 K
- V2 = ?

`(p_1 V_1)/(T_1) = (p_2 V_2)/(T_2)`

`(p_1 V_1)/(T_1) * (T_2)/(p_2) = V_2`

`(305 kPa* 1.0 L)/(403.15 K) * (366.15 K)/(101kPa) = V_2`

`V_2 = 2.74 L -> answer`

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