Un recipiente contiene 224 dm3 de Ozono de masa 4.561 Kg a 51.09 grados celsius. Calcula la presión del Ozono
1 answer:
Answer:
Por lo tanto, la presión del ozono es:
Explanation:
Podemos usar la ecuacion de los gases ideales;
(1)
Tenemos:
El volumen V = 224 dm³ = 224 L
La temperatura T = 51.09 C = 324.09 K
La masa es m = 4.561 kg
Lo necesitamos ahora es calvular n que es el numero de moles;
recordemos que el peso molecular del ozono M = 48 g/mol.
Finalmente, usando la ecuacion 1 despejamos la presion P
Por lo tanto, la presion del ozono es:
Espero te haya ayudado!
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Answer:
Hi
Final temperature = 250.11 °C
Final volume = 0,1 m3.
Process work = 0
Explanation:
The specific volume in the initial state is: v = 0.1m3/2 kg = 0.05 m3/kg.
This volume is located between the volumes as saturated liquid and saturated steam at 20 °C. For this reason the water is initially in a liquid vapor mixture. As the piston was blocked the volume remains constant and the process is isometric, also known as isocoric process, so the final temperature will be the water temperature at a saturated steam of v=0.05m3/kg, which is obtained by using steam tables for water, by linear interpolation. As follows, using table A-4 of the Cengel book 7th Edition:
v=0.05 m3/kg
v1=0.057061 m3/kg
T1=242.56°C
v2=0.049779 m3/kg
T2=250.35°C
T=
The process work is zero because there is no change in volume during heating:
W=PxΔv=Px0=0
where
W=process work
P=pressure
Δv=change of volume, is zero because the piston was blocked so the volume remains constant.