Question 1

Why does internal energy of a perfect gas depend only on temperature?

Accepted Answer

A perfect gas obeys pV = nRT, and its internal energy is unaffected by volume changes. Since pressure and volume are related through temperature in the equation of state, internal energy depends solely on temperature, not on how the gas is compressed or expanded. This fundamental property simplifies thermodynamic calculations for ideal gases.

Question 2

What happens to heat and work during an isothermal expansion of an ideal gas?

Accepted Answer

In an isothermal process, temperature remains constant, so the change in internal energy equals zero. According to the first law, heat absorbed equals the work done by the gas. During reversible isothermal expansion, the gas performs work on surroundings while absorbing an equal amount of heat from the constant-temperature bath.

Question 3

How does the equation of state relate to work calculations in reversible isothermal expansion?

Accepted Answer

The equation of state pV = nRT allows pressure to be expressed as p = nRT/V. Since temperature is constant during isothermal expansion, it can be factored outside the integral when calculating pressure-volume work. Integrating this expression relates the work done to the volume change from initial to final state.

Question 4

What is the key difference between adiabatic and isothermal expansion?

Accepted Answer

In isothermal expansion, temperature stays constant and heat flows into the gas to maintain internal energy. In adiabatic expansion, no heat is transferred, so the gas cools as it performs work. The internal energy decreases in adiabatic expansion, causing temperature to drop, whereas isothermal expansion maintains constant temperature throughout.

Question 5

How does adiabatic compression affect gas temperature and internal energy?

Accepted Answer

During adiabatic compression, no heat is transferred to or from the gas. Work done on the gas increases its internal energy, raising its temperature. The relationship between temperature and volume changes can be derived by equating work to internal energy change and substituting the ideal gas relation for pressure.

Question 6

Why does a constant-temperature bath maintain gas temperature during reversible isothermal expansion?

Accepted Answer

When external pressure decreases slowly, the gas expands and its temperature drops infinitesimally. This temperature decrease triggers heat flow from the bath into the gas, restoring the original temperature. This continuous heat transfer maintains constant temperature throughout the reversible isothermal process.

Question 7

How does the first law apply to adiabatic processes?

Accepted Answer

In adiabatic processes, no heat is transferred, so q = 0. The first law simplifies to ΔU = w, meaning the change in internal energy equals the work done. For adiabatic expansion, work is negative (gas does work), so internal energy decreases and temperature drops. For adiabatic compression, work is positive, increasing internal energy and temperature.

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