Animated Solution for Physics - Thermodynamics: Three moles of an ideal gas ($C_p=\frac{7}{2}R$) at pressure, $p_A$ and temperature $T_A$ is isothermally expanded to twice its initial volume. It is then compressed at constant pressure to its original volume. Finally gas is compressed at constant volume to its original pressure $p_A$. (a) Sketch $p-V$ and $p-T$ diagrams for the complete process. (b) Calculate the net work done by the gas, and net heat supplied to the gas during the complete process.

Imagine you are tracking the journey of three moles of an ideal gas through a complete thermodynamic cycle. We start at state $A$, where the gas has a pressure $p_A$, volume $V_A$, and temperature $T_A$. The gas undergoes three distinct processes:

1. Isothermal Expansion ($A\to B$): The gas expands to twice its initial volume ($V_B=2V_A$) while keeping the temperature constant ($T_B=T_A$). According to Boyle's Law ($p\propto 1/V$), doubling the volume halves the pressure. Thus, $p_B=p_A/2$. 2. Isobaric Compression ($B\to C$): The gas is compressed back to its original volume ($V_C=V_A$) at a constant pressure ($p_C=p_B=p_A/2$). According to Charles's Law ($V\propto T$), halving the volume halves the temperature. Thus, $T_C=T_A/2$. 3. Isochoric Compression ($C\to A$): The gas is heated at constant volume ($V_A$) until its pressure returns to $p_A$. According to Gay-Lussac's Law ($p\propto T$), doubling the pressure doubles the temperature, bringing it perfectly back to $T_A$.

To truly understand the cycle, we must sketch it on $p-V$ and $p-T$ diagrams.

On the $p-V$ diagram, the isothermal expansion ($A\to B$) is a curve (a rectangular hyperbola) moving downwards and to the right. The isobaric compression ($B\to C$) is a horizontal line moving to the left. Finally, the isochoric compression ($C\to A$) is a vertical line moving upwards, closing the clockwise cycle.

On the $p-T$ diagram, the isothermal process ($A\to B$) is a vertical line moving downwards (pressure drops at constant $T$). The isobaric process ($B\to C$) is a horizontal line moving to the left (temperature drops at constant $p$). The isochoric process ($C\to A$) is a straight line moving diagonally upwards. Because $pimesT$ at constant volume, this line, if extended, would pass perfectly through the origin!

Let's calculate the work done and heat exchanged during the first leg of the journey. For an isothermal process, the work done is given by:

Substituting $n=3$ and $V_B/V_A=2$:

Since the temperature is constant, the change in internal energy is zero ($\Delta U_{AB}=0$). By the First Law of Thermodynamics ($Q=W+\Delta U$), the heat supplied equals the work done:

Next, the gas is compressed at constant pressure. The heat exchanged is calculated using the molar heat capacity at constant pressure, $C_p=\frac{7}{2}R$:

Substituting $T_C=T_A/2$ and $T_B=T_A$:

The negative sign indicates that heat is released by the gas into the surroundings.

Finally, the gas is heated at constant volume. First, we need the molar heat capacity at constant volume, $C_V$. Using Mayer's relation ($C_p-C_V=R$):

Since the volume is constant, no work is done ($W_{CA}=0$). The heat supplied is:

Substituting $T_C=T_A/2$:

Net Work and Heat

We have completed the cycle! In any cyclic process, the initial and final states are identical, meaning the net change in internal energy is zero ($\Delta U_{\text{cycle}}=0$). Consequently, the net heat supplied must exactly equal the net work done ($Q_{\text{net}}=W_{\text{net}}$).

Let's sum up the heat from all three processes:

This elegant result confirms that the net heat supplied to the gas during the entire cycle is $0.58R{T}_A$, which is also the net work done by the gas.