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Explain why an ideal monatomic gas has no internal potential energy?
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The atoms in an ideal gas do not interact with each other via conservative forces. Because potential energy arises from interactions between objects, the absence of such forces means there is no internal potential energy. The internal energy consists entirely of the translational kinetic energies of the atoms.
A sealed container of ideal monatomic gas is heated, doubling the absolute temperature. Indicate whether the internal energy doubles, more than doubles, or less than doubles? Justify your answer.
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The internal energy exactly doubles. Internal energy is given by $U=\frac{3}{2}nRT$, which shows that $U$ is directly proportional to $T$. Since the container is sealed, $n$ is constant, so doubling $T$ doubles $U$.
A box of gas slides across a table and is brought to rest by friction. Explain how the internal energy of the gas can change even though the gas was not heated?
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Friction does work on the box, converting the kinetic energy of the center of mass into thermal energy. This increases the random kinetic energies of the gas molecules inside the box. The internal energy rises because energy is transferred into the system through work, not because of heating.
A gas in a sealed rigid container is cooled, losing 350 J of energy. Determine the change in internal energy of the gas.
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Because the container is rigid, $\Delta V=0$, so: $$W=-P\Delta V=0$$ Applying the first law: $$\Delta U=Q+W=-350+0$$ $$\Delta U=-350 \text{J}$$ The internal energy decreases by 350 J.
An ideal gas is compressed at a constant pressure of $2.0\times {10}^{5} \text{Pa}$, and its volume decreases by $3.0\times {10}^{-3} {\text{m}}^{3}$. During this process, 400 J of energy leaves the gas by cooling. Calculate the change in internal energy.
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Equation used: $$W=-P\Delta V$$ Given: $$P=2.0\times {10}^{5} \text{Pa}$$ $$\Delta V=-3.0\times {10}^{-3} {\text{m}}^{3}$$ Working: $$W=-(2.0\times {10}^{5})(-3.0\times {10}^{-3})$$ $$W=+600 \text{J}$$ Equation used: $$\Delta U=Q+W$$ Given: $$Q=-400 \text{J}$$ $$W=+600 \text{J}$$ Working: $$\Delta U=-400+600$$ $$\Delta U=+200 \text{J}$$ The internal energy increases by 200 J. The compression adds 600 J, but 400 J leaves as heat, giving a net gain.
Indicate whether the internal energy of an ideal gas increases, decreases, or stays the same when the gas expands and no heating or cooling occurs? Justify your answer.
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The internal energy decreases. With no heating or cooling, $Q=0$, so $\Delta U=W$. During expansion, $\Delta V>0$, which makes $W=-P\Delta V$ negative. A negative $\Delta U$ means the gas loses internal energy.
Two processes connect the same initial and final states on a PV diagram, but one follows a straight line and the other follows a curve that bows outward. Indicate which process involves more work done by the gas? Justify your answer.
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The process following the outward-bowing curve involves more work done by the gas. The magnitude of work equals the area under the curve on a PV diagram. The outward-bowing curve encloses a larger area beneath it than the straight line, so more energy is transferred as work.
Explain how you can determine from a PV diagram whether the temperature of an ideal gas increases during a process?
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For an ideal gas, $T=\frac{PV}{nR}$, so temperature is proportional to the product $PV$. If the process moves to a state where $PV$ is larger, the temperature increases. Graphically, higher isotherms (further from the origin) correspond to higher temperatures, so a process crossing to a higher isotherm indicates a temperature increase.
An ideal gas undergoes an isothermal expansion. Indicate whether the heat transferred to the gas is positive, negative, or zero? Justify your answer.
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The heat transferred is positive. During an isothermal process, $\Delta U=0$ because temperature is constant. The first law gives $Q=-W$. During expansion, $W=-P\Delta V$ is negative, so $Q=-W$ is positive. Energy must flow into the gas by heating to keep the temperature constant while the gas does work.
An ideal gas is compressed adiabatically. Predict what happens to the temperature of the gas? Justify using the first law.
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The temperature increases. In an adiabatic process, $Q=0$, so $\Delta U=W$. Compression means $\Delta V<0$, making $W=-P\Delta V$ positive. A positive $\Delta U$ means the internal energy increases. Because internal energy of an ideal gas is proportional to temperature, the temperature rises.
Derive an expression for the heat transferred to an ideal monatomic gas during an isobaric expansion in terms of $P$, $\Delta V$, and fundamental constants.
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Starting point: The first law of thermodynamics. $$\Delta U=Q+W$$ Step 1 — Rearrange to isolate $Q$. $$Q=\Delta U-W$$ Step 2 — Substitute the expression for work at constant pressure. $$W=-P\Delta V$$ $$Q=\Delta U-(-P\Delta V)=\Delta U+P\Delta V$$ Result: $$Q=\Delta U+P\Delta V$$ For a monatomic ideal gas, $\Delta U=\frac{3}{2}nR\Delta T$. Using the ideal gas law at constant pressure, $P\Delta V=nR\Delta T$, so $\Delta U=\frac{3}{2}P\Delta V$. Substituting: $$Q=\frac{3}{2}P\Delta V+P\Delta V=\frac{5}{2}P\Delta V$$ This shows that during an isobaric expansion, the gas requires more heat than the work it performs because internal energy must also increase.
Indicate whether the expression $Q=\frac{5}{2}P\Delta V$ is consistent with the claim that more energy enters the gas as heat than leaves as work during an isobaric expansion? Justify your answer.
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The expression is consistent. The work done by the gas during an isobaric expansion has magnitude $P\Delta V$. The heat transferred in is $\frac{5}{2}P\Delta V$. Since $\frac{5}{2}P\Delta V>P\Delta V$ for any positive $\Delta V$, more energy enters as heat than leaves as work. The extra energy increases the internal energy of the gas, which is consistent with the temperature rising during isobaric expansion.