Un cilindro de gas a 32 °C normalmente tiene una presión interna de 11 atm, sin embargo, la válvula de alivio se debería abrir antes de llegar a 19, 380 mmHg. ¿A qué temperatura es cuando la válvula se abre para liberar la presión interna?

T₂ = 72ºC

Explanation:

In this exercise let us reduce the pressure to the SI system

     P = 11 ata (1 10⁵ Pa / 1 atm) = 11 10⁵ P

     y = 19380 mmHg

pressure is

     P = ρ g h

     P = 13.6 10³  9.8   19380 10⁻³

     P = 2.5 10⁶ Pa

Now we can use the ideal gas equation to

        PV = nR T

        PV / T = n R

Let's use the subscript 1 for the starting point and the subscript 2 for the point where the valve opens at a constant volume

         P₁ V / T₁ = P₂ V / T₂

         T₂ = P₂ T₁/ P₁

let's calculate

          T₂ = (2.5 10⁶/11 10⁵ 32

          T₂ = 72ºC

change tension

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )