Question

An ideal gas undergoes a constant pressure (isobaric) process in a closed system. The heat transfer and work are, respectively \((a) 0,-c_{v} \Delta T\) (b) \(c_{v} \Delta T, 0\) \((c) c_{p} \Delta T, R \Delta T\) \((d) R \ln \left(T_{2} / T_{1}\right), R \ln \left(T_{2} / T_{1}\right)\)

Short answer

Question: An ideal gas undergoes an isobaric process in a closed system. Derive expressions for the work done (W) and the heat transfer (Q) during this process. Answer: For an isobaric process involving an ideal gas, the work done (W) and the heat transfer (Q) are given by W = \(R \Delta T\) and Q = \(c_p \Delta T\), respectively.

Step-by-step solution

Define the given variables and the goal

In this problem, we are given an ideal gas that undergoes an isobaric process. We are asked to find the heat transfer (Q) and work (W) for this process. The given options contain variables \(c_v\), \(c_p\), \(R\), \(\Delta T\), \(T_1\), and \(T_2\). These represent heat capacities at constant volume and pressure, the gas constant, the temperature change, and the initial and final temperatures, respectively.

Ideal gas law and initial and final states

For an ideal gas, the ideal gas law is given by: \(PV=nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Since the process is isobaric, we can write the ideal gas law for the initial and final states as: \(P_1V_1 = nRT_1\) \(P_2V_2 = nRT_2\) Since the pressure is constant throughout the process, we have: \(PV_1 = nRT_1\) \(PV_2 = nRT_2\)

Work done during an isobaric process

In an isobaric process, the work done can be calculated as: \(W = P \Delta V = P(V_2 - V_1)\) Using the ideal gas law from Step 2, we can rewrite the work expression as: \(W = P(V_2 - V_1) = R \Delta T\)

First law of thermodynamics and heat transfer

The first law of thermodynamics can be written as: \(\Delta U = Q - W\) For an ideal gas, the change in internal energy only depends on the heat capacities at constant volume (\(c_v\)) and the temperature change: \(\Delta U = n c_v \Delta T\) So, we can write the first law of thermodynamics as: \(nc_v \Delta T = Q - R \Delta T\) Now, let's solve for Q: \(Q = (c_v + R) \Delta T = c_p \Delta T \) Here, \(c_p\) is the heat capacity at constant pressure. This is because, for an ideal gas, \(c_p = c_v + R\). Thus, the heat transfer is \(c_p \Delta T\).

Compare expressions to given options

Now we can compare our expressions derived in Steps 3 and 4 for the work done (W) and the heat transfer (Q) to the given options: W = \(R \Delta T\) Q = \(c_p \Delta T\) Comparing these expressions, we see that they match option (c): \((c) c_{p} \Delta T, R \Delta T\) So, the correct answer is \((c) c_{p} \Delta T, R \Delta T\).