Analyzing Thermodynamic Process Relations for Work and Heat

The question asks us to identify the INCORRECT relation among the given options for different thermodynamic processes concerning work done (W) and heat transfer (Q). Let's examine each option based on the definitions and the First Law of Thermodynamics (\(Q = \Delta U + W\)) for a closed system, assuming ideal gas behavior where appropriate.

Option 1: Isobaric process

An isobaric process occurs at constant pressure (\(p\)).

• Work done (W): For a constant pressure process, the work done is given by the integral of pressure with respect to volume change. \(W = \int_{V_1}^{V_2} p \, dV\). Since \(p\) is constant, \(W = p \int_{V_1}^{V_2} dV = p(V_2 - V_1)\). In terms of specific volume (\(v\)), \(W = p(v_2 - v_1)\). This part of the relation is correct.
• Heat transfer (Q): According to the First Law, \(Q = \Delta U + W\). For an ideal gas, \(\Delta U = m c_v (T_2 - T_1)\) or \(\Delta U = c_v (T_2 - T_1)\) per unit mass. For an isobaric process, $W = p(v_2 - v_1) = R(T_2 - T_1)$ (using $pv=RT$). Thus, \(Q = c_v(T_2 - T_1) + R(T_2 - T_1) = (c_v + R)(T_2 - T_1)\). Since \(c_p = c_v + R\) for an ideal gas, \(Q = c_p(T_2 - T_1)\). This part of the relation is also correct.

So, the relations \(W = p(v_2 - v_1)\) and \(Q = c_p(T_2 - T_1)\) for an isobaric process are CORRECT.

Option 2: Isothermal Process

An isothermal process occurs at constant temperature (\(T\)).

• Work done (W): For an isothermal process involving an ideal gas, \(pv = RT = \text{constant}\). \(W = \int_{V_1}^{V_2} p \, dV = \int_{V_1}^{V_2} \frac{RT}{V} \, dV = RT \int_{V_1}^{V_2} \frac{1}{V} \, dV = RT [\ln V]_{V_1}^{V_2} = RT \ln\left(\frac{V_2}{V_1}\right)\). Using \(p_1V_1 = RT\), this can also be written as \(W = p_1V_1 \ln\left(\frac{V_2}{V_1}\right)\) or \(W = p_1v_1 \ln\left(\frac{v_2}{v_1}\right)\) per unit mass. For an isothermal process, work done is generally NOT zero unless there is no volume change. The option states \(W = 0\), which is incorrect for typical isothermal expansion or compression.
• Heat transfer (Q): For an ideal gas, internal energy \(U\) depends only on temperature. Since the temperature is constant (\(T_1 = T_2\)), the change in internal energy \(\Delta U = c_v(T_2 - T_1) = 0\). According to the First Law, \(Q = \Delta U + W = 0 + W = W\). So, \(Q = W = p_1v_1 \ln\left(\frac{v_2}{v_1}\right)\). The option states \(Q = p_1v_1 \log_e\left(\frac{T_1}{T_2}\right)\). Since \(T_1 = T_2\) in an isothermal process, \(\frac{T_1}{T_2} = 1\), and \(\log_e(1) = 0\). This formula would incorrectly give \(Q=0\).

Therefore, the relations \(W = 0\) and \(Q = p_1v_1 \log_e\left(\frac{T_1}{T_2}\right)\) for an isothermal process are INCORRECT.

Option 3: Isochoric process

An isochoric process occurs at constant volume (\(V\) or \(v\)).

• Work done (W): \(W = \int_{V_1}^{V_2} p \, dV\). Since \(V_1 = V_2\), \(dV = 0\). Thus, \(W = 0\). This part of the relation is correct.
• Heat transfer (Q): According to the First Law, \(Q = \Delta U + W\). Since \(W=0\), \(Q = \Delta U\). For an ideal gas, \(\Delta U = c_v(T_2 - T_1)\). So, \(Q = c_v(T_2 - T_1)\). This part of the relation is correct.

So, the relations \(W = 0\) and \(Q = c_v(T_2 - T_1)\) for an isochoric process are CORRECT.

Option 4: Throttling process

A throttling process is typically considered an adiabatic (\(Q=0\)) and irreversible process where a fluid flows through a restriction (like a valve) with no external work done (\(W=0\)) and negligible changes in kinetic and potential energy. This leads to an isenthalpic process (\(\Delta h = 0\)).

• Work done (W): In the context of boundary work for a closed system, throttling does not involve volume change against a boundary, so \(W=0\). For a steady-flow open system analysis with no shaft work, the concept of work done is different, but for typical questions like this, W is often considered zero external work. This part of the relation is generally considered correct in this context.
• Heat transfer (Q): Throttling is usually assumed to be adiabatic, meaning no heat transfer occurs, so \(Q = 0\). This part of the relation is correct.

So, the relations \(W = 0\) and \(Q = 0\) for a throttling process are generally considered CORRECT under typical assumptions.

Based on the analysis, the INCORRECT option is the one describing the Isothermal Process relations.