Answer:
a) 
b) entropy of the sistem equal to a), entropy of the universe grater than a).
Explanation:
a) The change of entropy for a reversible process:
The energy balance:
![\delta U=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=%5Cdelta%20U%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)
If the process is isothermical the U doesn't change:
![0=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=0%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)
The work:
If it is an ideal gas:
Solving:
Replacing:
Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:
b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.
Gases and plasma's undergo changes in volume most easily<span>. In contrast, liquids and solids have fixed volumes, although liquids do not have fixed shapes. Gases and plasma's expand or contract to fill their containers.</span><span />
7+8=15 es un cation ok ok ok ok
Answer:
443 L of carbon dioxide
Explanation:
Combustion reaction: CH₄ (g) + 2O₂ (g) → CO₂(g) + 2H₂O(g)
We assume, that the oxygen is the excess reagent, so let's convert the mass of methane to moles. (mass (g) / molar mass)
Firstly we need to convert the mass from kg to g → 0.300 kg .1000 g / 1kg
300 g / 16 g/mol = 18.75 moles of methane.
Ratio is 1:1, so 18.75 moles of methane will produce 18.75 moles of CO₂
We apply the Ideal Gases Law to find the answer:
Firstly we need to convert the T°C to T°K → 15°C + 273 = 288 K
P . V = n . R. T → V = ( n . R . T ) / P We replace data:
V = (18.75 mol . 0.0821 L.atm/mol.K . 288K) / 1 atm → 442.8 ≅ 443 L
