Respuesta :
Answer:
71.19 C
Explanation:
25C = 25 + 273 = 298 K
Applying the ideal gas equation we have
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:
[tex]T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C[/tex]
The ideal gas equation for the two points allows finding the gas temperature when the balloon is elevated is 71.2ºC
Ideal gases is a very good primer for gases where interactions between them are not taken into account, the expression is
P V = n R T
Where P is the pressure, V the volume, n the number of moles, R the ideal gas constant and T the absolute temperature.
Let's write this ideal gases equation for the two points, let's use the subscript 1 for when it is on the ground and the subscript 2 for when the balloon is raised.
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
T2 = [tex]\frac{P_2}{P1} \frac{V_2}{V_1} \ T_1[/tex]
The international system of measurements is a system of measurements that allows the exchange of quantities and measurements without inconvenience and in the correct way. Let us reduce the quantities to the SI system.
T₁ = 25 +273 = 298K
We can see in the equation that the pressures and the volumes are divided by which the reduction factor is eliminated, we can leave them without reducing.
let's calculate
T2 = [tex]\frac{0.77}{1.0} \frac{1.8}{1.2} \ 298[/tex]
T2 = 344.2K
Let's reduce the temperature
T₂ = 2344.2 - 273
T₂ = 71.2ºC
Using the ideal gas equation for the two points we can calculate the gas temperature when the balloon is elevated is 71.2ºC
Learn more here: brainly.com/question/24049206
