A metal bar increases in length by 0.05% when heated from 0°C to 100°C. Calculate its coefficient of linear expansion and the percentage increase in its volume.

The temperature change is $\Delta T = 100^\circ\text{C} - 0^\circ\text{C} = 100^\circ\text{C}$.
The percentage increase in length is $0.05\%$, so the fractional increase in length is $\frac{\Delta L}{L_0} = \frac{0.05}{100} = 0.0005$.

The formula for linear expansion is $\Delta L = \alpha L_0 \Delta T$. Rearranging for $\alpha$:

The coefficient of linear expansion is:

Part 15b(ii): Percentage increase of the volume of the bar

For isotropic materials, the coefficient of volume expansion is approximately three times the coefficient of linear expansion:

The fractional increase in volume is $\frac{\Delta V}{V_0} = \gamma \Delta T$.

To express this as a percentage:

The percentage increase of the volume of the bar is:

Part 16a(i): Quantity that remains constant for thermodynamic processes

• A) Isobaric process: Pressure ($P$)
• B) Isochoric process: Volume ($V$)
• C) Isothermal process: Temperature ($T$)
• D) Adiabatic process: Heat exchange ($Q=0$)

Part 16a(ii): Show that $PV^\gamma = \text{constant}$ for a monatomic ideal gas undergoing an adiabatic process

For an adiabatic process, there is no heat exchange, so $dQ = 0$.
The First Law of Thermodynamics states $dQ = dU + dW$.
Therefore, $dU + dW = 0$.

For an ideal gas, the change in internal energy is $dU = nC_v dT$, where $n$ is the number of moles and $C_v$ is the molar specific heat at constant volume.
The work done by the gas is $dW = P dV$.
Substituting these into the equation from Step 1:

The ideal gas law is $PV = nRT$. Differentiating both sides:

Rearranging for $dT$:

Substitute (2) into (1):

We know that $R = C_p - C_v$, where $C_p$ is the molar specific heat at constant pressure. So, $\frac{C_v}{C_p - C_v} (P dV + V dP) + P dV = 0$.
Divide by $C_v$:

Let $\gamma = \frac{C_p}{C_v}$ (the adiabatic index).

Multiply by $(\gamma - 1)$:

Divide the equation by $PV$:

Integrate both sides:

Exponentiating both sides:

Since $e^{\text{constant}}$ is also a constant, we have:

Part 16b(i): Efficiency for a Carnot cycle

A) As a function of the heat that flows to and from the reservoirs:

where $Q_C$ is the heat rejected to the cold reservoir and $Q_H$ is the heat absorbed from the hot reservoir.

B) As a function of the temperatures of the two reservoirs: