During an isothermal compression of an ideal gas, $410$ J of heat must be removed from the gas to maintain constant temperature. How much work is done by the gas during the process?

A cylinder contains $0.0100$ mol of helium at $T = 27.0$°C. If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from $27.0$°C to $67.0$°C? Draw a $pV$-diagram for this process.

A cylinder contains $0.0100$ mol of helium at $T = 27.0$°C. What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat?

(a) How much heat is needed to raise the temperature to $67.0$°C while keeping the volume constant? Draw a $pV$-diagram for this process.

(b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from $27.0$°C to $67.0$°C? Draw a $pV$-diagram for this process.

When water is boiled at a pressure of $2.00$ atm, the heat of vaporization is $2.20\(\times\)10^6$ J/kg and the boiling point is $120$°C. At this pressure, $1.00$ kg of water has a volume of $1.00\(\times\)10^{-3}$ m³, and $1.00$ kg of steam has a volume of $0.824$ m³. Compute the increase in internal energy of the water.

An ideal gas is taken from $a$ to $b$ on the $pV$-diagram shown in Fig. E$19.15$. During this process, $700$ J of heat is added and the pressure doubles. How does the internal energy of the gas at $a$ compare to the internal energy at $b$? Be specific and explain.

A cylinder contains $0.0100$ mol of helium at $T=27.0$°C. How much heat is needed to raise the temperature to $67.0$°C while keeping the volume constant? Draw a $pV$-diagram for this process.

When water is boiled at a pressure of $2.00$ atm, the heat of vaporization is $2.20\times {10}^6$ J/kg and the boiling point is $120$°C. At this pressure, $1.00$ kg of water has a volume of $1.00\times {10}^{-3}$ m³, and $1.00$ kg of steam has a volume of $0.824$ m³. Compute the work done when $1.00$ kg of steam is formed at this temperature.