Answer:
the entropy change of the fluid during the process process is is 1.337 kJ/K, the change for the source is -1.337 kJ/K and the total entropy change is 0
Explanation:
since the Carnot cycle is a reversible cycle, the entropy change is related with the heat exchanged through:
ΔS =∫dQ/T
since the temperature remains constant
ΔS =∫dQ/T=(1/T)*∫dQ = Q/T
Q= heat added to the system
T= absolute temperature = 400°C= 673 K
therefore
ΔS = Q/T = 900 kJ/ 673 K = 1.337 kJ/K
ΔS working fluid = 1.337 kJ/K
since the process is reversible, the entropy change of the universe (total entropy change) is 0 (there is no entropy generation). thus
ΔS universe = ΔS working fluid + ΔS source = 0
ΔS source= -ΔS working fluid = -1.337 kJ/K
Answer: The ratio of the number of oxygen molecules to the number of nitrogen molecules in these flasks is 1: 1
Explanation:
According to avogadro's law, equal volumes of all gases at same temperature and pressure have equal number of moles.
According to avogadro's law, 1 mole of every substance contains avogadro's number of particles.
Thus as oxygen and nitrogen are at same temperature and pressure and are in equal volume flasks , they have same number of moles and thus have same number of molecules.
The ratio of the number of oxygen molecules to the number of nitrogen molecules in these flasks is 1: 1
The equation is:4Al+3O2----2Al2O3
The final volume of the gas that was heated from -25.0 °C to standard temperature is 2.2L.
<h3>How to calculate volume?</h3>
The volume of a given gas can be calculated using the Charles law equation as follows:
V1/T1 = V2/T2
Where;
- V1 = initial volume
- V2 = final volume
- T1 = initial temperature
- T2 = final temperature
- V1 = 2L
- V2 = ?
- T1 = -25°C + 273 = 248K
- T2 = 273K
2/248 = V2/273
273 × 2 = 248V2
546 = 248V2
V2 = 546/248
V2 = 2.2L
Therefore, the final volume of the gas that was heated from -25.0 °C to standard temperature is 2.2L
Learn more about volume at: brainly.com/question/11464844