i need the answer asap plz!!
A sample of water vapor has a pressure of 36.0 Pa and a temperature of 300.0 K. If the pressure is
reduced to 32.0 Pa, what will the new temperature be?

Respuesta :

Answer:

266.7k

Explanation:

Given parameters:

Initial pressure  = 36Pa

Initial temperature  = 300k

New pressure  = 32Pa

Unknown:

New temperature  = ?

Solution:

To solve this problem, we have to apply the combined gas law when the volume is constant;

in this instant, when the volume is constant, the pressure of a given mass of gas varies directly with the absolute temperature.

          [tex]\frac{P_{1} }{T_{1} }[/tex]   = [tex]\frac{P_{2} }{T_{2} }[/tex]  

P and T are pressure and temperature

1 and 2 are initial and final states

  Insert the parameters and solve;

       [tex]\frac{36}{300}[/tex]   = [tex]\frac{32}{T_{2} }[/tex]  

      T₂  = 266.7k

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