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Arturiano [62]
3 years ago
6

The pressure of a gas is 1.0 atm, the volume is 3.0 L, and the temperature is 200 K. A chemist changes one factor while keeping

another constant so that the new pressure is 2.0 atm. Which of the following could be the new conditions?
The final volume is 6.0 L, while temperature is kept constant.
The final temperature is 100 K, while volume is kept constant.
The final volume is 4.0 L, while temperature is kept constant.
The final temperature is 400 K, while volume is kept constant.
Chemistry
1 answer:
anygoal [31]3 years ago
4 0

Answer: fourth statement is correct. The final temperature is 400 K, while volume is kept constant.

Explanation: At constant temperature, volume is inversely proportional to the pressure and,

At constant volume, pressure is directly proportional to the kelvin temperature.

Initially the gas pressure is 1.0 atm, volume is 3.0 L and the temperature is 200 K.  The new pressure is 2.0 atm.

If we consider the temperature is constant, then new volume must be 1.5 L since the pressure is doubled and we know that volume is inversely proportional to the pressure.

So, as the pressure is doubled, the volume will be halved means it will decrease from 3.0 L to 1.5 L.

First statement is not correct since it says the new volume will be 6.0 L while temperature is kept constant.

If the volume is kept constant, the temperature must be doubled as the pressure has also doubled. So, new temperature will be 400 K.

second statement is also not correct as it says, the final temperature is 100 K, while volume is kept constant.

Third statement is also not correct as we already have seen that the new volume will be 1.5 L, while the temperature is kept constant.

Hence, the only and only correct statement is fourth as the final temperature will be 400 K(double of initial 200 K), while the volume is kept constant.

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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:

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The energy balance:

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

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The work:

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

If it is an ideal gas:

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

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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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