All of the following equations are statements of the ideal gas law except
1 answer:
Answer:
- The first equation, <em>a. PV = nRT</em>, <u>is not</u> <em>a valid statement of the ideal gas law.</em>
Explanation:
The basic expression for the<em> ideal gas law</em> is:
.......... [Equation 1]
Where:
- n is the number of moles of the gas
- V is the volume occupied by the gas
- p is the pressure exerted by the gas molecules
- T is the temperature in absolute scale (Kelvin)
- R is the Universal gas constant (0.0821 atm-liter /K-mol or the equivalents in other units)
You can perform different algebraic operations to obtain equivalent equations:
<u>Choice b) Divide equation 1 by T and you get</u>:
- pV / T = nR, which is the choice b. from your list.
<u>Choice c) Divide equation 1 by n × V and you get</u>:
- p/n = RT / V, which is the choice c. from your list.
<u>Choice d) Divide equation 1 n × T and you get</u>:
- pV / (nT) = R, which is the choice d. from your list.
The choice a. p = nRTV states that p and V are in direct relation, when the ideal gas law states that p and V are inversely related, so that equation is wrong.
<u>Conclusion: </u>the choice a, p = nRTV, is not a statement of the ideal gas law.
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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:
If the process is isothermical the U doesn't change:
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.