Answer:
es un motor de combustión interna con encendido por chispa.
The temperature of the gas is 41.3 °C.
Answer:
The temperature of the gas is 41.3 °C.
Explanation:
So on combining the Boyle's and Charles law, we get the ideal law of gas that is PV=nRT. Here P is the pressure, V is the volume, n is the number of moles, R is gas constant and T is the temperature. The SI unit of pressure is atm. So we need to convert 1 Pa to 1 atm, that is 1 Pa = 9.86923×
atm. Thus, 171000 Pa = 1.6876 atm.
We know that the gas constant R = 0.0821 atmLMol–¹K-¹. Then the volume of the gas is given as 50 L and moles are given as 3.27 moles.
Then substituting all the values in ideal gas equation ,we get
1.6876×50=3.27×0.0821×T
Temperature = 
So the temperature is obtained to be 314.3 K. As 0°C = 273 K,
Then 314.3 K = 314.3-273 °C=41.3 °C.
Thus, the temperature is 41.3 °C.
Answer:
57,42 KJ
Explanation:
By a isobaric proces, the expresion for the works in the jpg adjunt. Then:
W = Pa(Vb - Va) = Pa*Vb - Pa*Va ---(1)
By the ideal gases law: PV=RTn
Then, in (1): (remember Pa = Pb)
W = R*Tb*n - R*T*an = R*n*(Tb - Ta) --- (2)
Since we have 1 Kg air: How much is this in moles?
From bibliography: 28.96 g/mol
Then, in 1 Kg (1000 g) there are:
n = 34,53 mol
Finally, in (2):
W = (8,3144 J/K.mol)*(34,53 mol)*(500K - 300K) = 51 419,9 J ≈ 57,42 KJ