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. 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.

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?

The $pV$-diagram in Fig. E$19.13$ shows a process $abc$ involving $0.450$ mol of an ideal gas. How much heat had to be added during the process to increase the internal energy of the gas by $15,000$ J?

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.

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.