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Likurg_2 [28]
3 years ago
10

Why is basalt a sign that the moon once had magma at its surface14 points

Chemistry
1 answer:
lawyer [7]3 years ago
7 0

The basalt is a sign that the Moon once had magma at its surface because the basalt is a type of rock that forms directly from the cooling off of the magma. The basalt is part of the igneous type of rocks, which are the type of rocks that form from the cooling of the magma, be it on the surface or inside the ground, so having basalt on the Moon is a sure sign that our natural satellite once had magma on it.

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An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
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Answer:

a) \Delta S

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:

\delta S=\frac{\delta Q}{T}

\Delta S=\frac{Q}{T}

The energy balance:

\delta U=[tex]\delta Q- \delta W

If the process is isothermical the U doesn't change:

0=[tex]\delta Q- \delta W

\delta Q= \delta W

Q= W

The work:

W=\int_{V1}^{V2}P*dV

If it is an ideal gas:

P=\frac{n*R*T}{V}

W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV

Solving:

W=n*R*T*ln(V2/V1)

Replacing:

\Delta S=\frac{n*R*T*ln(V2/V1)}{T}

\Delta S=n*R*ln(V2/V1)}

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

\Delta S

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.

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Answer:

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Explanation:

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