# Step-by-Step Solution Let's clarify and answer each part of the problem about 1 kg of an ideal (dual) gas. --- ## (a) **Establishing a Reversible Process to Determine the Specific Heat at Constant Volume ($c_v$)** ### **Experiment:** - Place 1 kg of the ideal gas in a rigid (constant volume) and insulated cylinder. - Insert a thermometer and a heater with known electrical power. - Supply a known amount of heat ($Q$) to the gas. - Measure the temperature change ($\Delta T$). ### **Calculation:** - Since the volume is constant, all heat supplied goes into raising the internal energy: \[ Q = m c_v \Delta T \] where: - $Q$ is heat supplied (in joules, J) - $m$ is the mass of the gas (1 kg here) - $c_v$ is the specific heat at constant volume (J/kg·K) - $\Delta T$ is the temperature change (K) - Rearranging: \[ c_v = \frac{Q}{m \Delta T} \] ### **Units:** - $c_v$ will be in **J/kg·K**. --- ## (b) **Establishing a Reversible Process to Determine the Specific Heat at Constant Pressure ($c_p$)** ### **Experiment:** - Place 1 kg of the ideal gas in a cylinder with a light, frictionless, movable piston (constant pressure). - Supply a known amount of heat ($Q'$) to the gas. - Allow the gas to expand at the same pressure as it is heated. - Measure the temperature change ($\Delta T'$). ### **Calculation:** - At constant pressure, the heat supplied increases both the internal energy and does work as the gas expands: \[ Q' = m c_p \Delta T' \] - $c_p$ is the specific heat at constant pressure (J/kg·K) - Rearranging: \[ c_p = \frac{Q'}{m \Delta T'} \] ### **Units:** - $c_p$ will be in **J/kg·K**. --- ## (c) **First Law of Thermodynamics Relation** The **First Law of Thermodynamics** states: \[ \Delta U = Q - W \] - At **constant volume** ($W = $): $Q = \Delta U$ - At **constant pressure**: $Q = \Delta U + P\Delta V$ For an ideal gas: - $\Delta U = m c_v \Delta T$ - $P\Delta V = mR\Delta T$ (from $PV = mRT$, so $\Delta V = mR\Delta T / P$) So at constant pressure: \[ Q = m c_v \Delta T + mR \Delta T = m (c_v + R)\Delta T \] But at constant pressure, $Q = m c_p \Delta T$, so: \[ m c_p \Delta T = m (c_v + R)\Delta T \] \[ c_p = c_v + R \] where $R$ is the **specific gas constant** for the gas (in J/kg·K). --- ## **Summary and Final Answers** 1. **To determine $c_v$:** - Perform a reversible, constant volume heating experiment. - $c_v = \dfrac{Q}{m \Delta T}$ 2. **To determine $c_p$:** - Perform a reversible, constant pressure heating experiment. - $c_p = \dfrac{Q'}{m \Delta T'}$ 3. **First Law Proof:** - For an ideal gas: $\boxed{c_p = c_v + R}$ - $R$ is the specific gas constant (J/kg·K) of the given gas. --- ### **Summary Table** | Property | Symbol | Formula | Units | |----------------------|----------|--------------------------------|------------| | Specific heat (const. volume) | $c_v$ | $c_v = \frac{Q}{m\Delta T}$ | J/kg·K | | Specific heat (const. pressure) | $c_p$ | $c_p = \frac{Q'}{m\Delta T'}$ | J/kg·K | | Specific gas constant | $R$ | $c_p = c_v + R$ | J/kg·K | --- **Note:** - All measurements should ensure the process is slow (quasi-static) to be reversible. - The gas should behave ideally for these relations to hold.