Carnot’s Cycle

Carnot’s Cycle

Consider n-mole of an ideal gas enclosed in the cylinder. Let P₁, V₁, T₁ be the initial pressure, volume and temperature of the gas in it. The initial state of the gas is represented by the point A or the P-V diagram.

1. Isothermal Expansion
   First, the cylinder containing the working the substance is placed on the source and the gas is allowed to expend by slow outward motion of the piston. As the gas draws heat from the source through the conducting base of the cylinder, the gas expands at constant temperature T₁. This operation is called isothermal expansion and is represented by the isothermal line AB in figure. Let the quantity of heat absorbed in this process be Q₁ and W₁ be the corresponding amount of work done by the gas in expanding isothermally from A(P₁, V₁) to B(P₂V₂). Then according to 1^st law of thermodynamics.
   \begin{align*} Q_1 &= W_1 \\ &= \int _{v_1}^{v_2} P dV \\ &= nRT_1 \int _{v_1}^{v_2} \frac {dV}{V} \\ &= nRT_1 \log \frac {V_2}{V_1} \dots (i) \\ \end{align*}
2. Adiabatic Expansion
   The cylinder is removed from the source and then placed on the non-conducting stand. The gas is allowed to expand further from B (P₂, V₂) to C (P₃, V₃). Since the working substance is completely thermally insulated from all sides of the cylinder, no heat is gained from the surroundings. Therefore, as the gas is allowed to expand further, the temperature of the gas falls to T₂. The expansion is adiabatically in nature and the line BC represents the adiabatic expansion. Let W₂ be the work done by the gas in expanding adiabatically from B (P₂, V₂) to C (P₃, V₃).
   \begin{align*} W_2 &= \int _{v_3}^{v_2} P dV \\ \end{align*}Since in adiabatic process,\begin{align*}PV^{\gamma} = \text {constant} = K, \\ W_2 \int _{v_3}^{v_2} \frac {dV}{V^{\gamma }} \\ &= \frac {KV_3^{1-\gamma} – KV_2^{1-\gamma}}{1 - \gamma } \\ &= \frac {P_3V_3 –P_2V_2}{1 - \gamma } \: [\because P_2V_2^\gamma = P_3V_3^\gamma = K] \\ &= \frac {nRT_2 – nRT_1}{1 - \gamma } \: [\because P_2V_2 = nRT_1, P_3V_3 = nRT_2] \\ &= \frac {nR(T_2 – T_1)}{1 - \gamma } \\ \therefore \: W_2 &= \frac {nR(T_2 – T_1)}{1 - \gamma } \dots (ii) \\ \end{align*}
   This work done is represented by the area BCHGB in PV-diagram.
3. Isothermal Compression
   The cylinder is now removed from the insulating stand and is placed on the heat sink at a temperature T₂. The gas is compressed slowly so that as the heat is developed in the gas, it easily flows to the sink through the conducting base of the cylinder at constant temperature T₂ of the gas. This process is isothermal and is represented by the isothermal curve CD. Let Q₂ be the amount of heat energy rejected to the sink and W₃ be the work done on the working substance during isothermal compression from a state C(P₃, V₃) to D(P₄, V₄). So,
   \begin{align*} Q_2 = W_3 = - \int _{V_4}^{V_3} P\: dV \\ \end{align*}The negative sign shows the isothermal compression \begin{align*}\therefore Q_2 = W_3 = \int _{V_3}^{V_4} R\: dV \\ &= nRT_2 \int _{V_3}^{V_4} \frac {dV}{V} \\ &= nRT_2 \log_e \frac {V_3}{V_4} \dots (iii) \\ &= \text {Area} \: CHFDC \\ \end{align*}
4. Adiabatic Compression
   The cylinder is again transferred to the insulating stand and the piston is moved downwards so that the working substance is compressed adiabatically along DA to its initial pressure P1 and the volume V₁. As the working substance is insulated from all sides, heat produced raises the temperature of the gas to T₁. Let W₄ be the amount of work done on the working substance in compressing it adiabatically from a state D (P₄, V₄) to A (P₁, V₁). So,
   \begin{align*} W_4 &= -\int _{v_1}^{v_4} P dV \\ &= \frac {nR(T_1 – T_2) }{(\gamma -1)} \dots (iv) \\ \end{align*}
   This is represented by the area DFEAD in PV-diagram.

Work done by the engine per cycle
In steps 1 and 2, total work done by the gas = W₁ + W₂

In steps 3 and 4, total work done on the gas = W₃ + W₄

\(\therefore\) Net work done by the gas in one complete cycle,\begin{align*}\: W = W_1 + W_2 + (-W_3) + (-W_4) \\ \text {From equations} \: (ii) \text {and} \: (iv), \text {we have} \\ W_2 &= W_4 \\ \therefore W = W_1 – W_3 \\ \text {or,} \: W &= Q_1 – Q_2 \dots (v) \\ \end{align*}So, the net work done by the engine is one complete cycle is \begin{align*}W = \text {Area} \: ABGEA + \text {Area} \: BCHGB - \text {Area} \: CHFDC - \text {Area} \: DFEAD \end{align*}$$= \text {Area} \: ABCDA$$
Hence in Carnot engine, net work done by the gas per cycle is numerically equal to the area of the loop ABCDA representing the cycle.