Question

Consider a process in which an ideal gas changes from state 1 to state 2 in such a way that its temperature changes from \(300 \mathrm{~K}\) to \(200 \mathrm{~K}\). (a) Describe how this change might be carried out while keeping the volume of the gas constant. (b) Describe how it might be carried out while keeping the pressure of the gas constant. (c) Does the change in \(\Delta E\) depend on the particular pathway taken to carry out this change of state? Explain.

Short answer

To carry out the change while keeping the volume constant, we should decrease the pressure of the gas to two-thirds of its initial value. To carry out the change while keeping the pressure constant, we should decrease the volume of the gas to two-thirds of its initial value. The change in internal energy (∆E) depends only on the change in temperature and not on the particular pathway taken to carry out the change of state, no matter if we keep the volume or the pressure constant.

Step-by-step solution

Part a: Changing state while keeping volume constant

To keep the volume constant, we have to consider the formula: \(PV = nRT\) Since the volume remains constant, the relationship between pressure, temperature, and the constant R and number of moles (n) can be expressed as: \(P_1/T_1 = P_2/T_2\) Given the initial temperature \(T_1 = 300 \mathrm{~K}\) and the final temperature \(T_2 = 200 \mathrm{~K}\), we can calculate the ratio of pressures \(P_1\) and \(P_2\): \(\frac{P_1}{300} = \frac{P_2}{200}\) Solving for \(P_2\), we get the relationship \(P_2 = \frac{2}{3} P_1\) So to change the state of the gas while keeping the volume constant, we should decrease the pressure of the gas to two-thirds of its initial value.

Part b: Changing state while keeping pressure constant

To keep the pressure constant, we have to think again about the ideal gas law \(PV = nRT\). Since the pressure remains constant, we can express the relationship between the initial and final states as follows: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) Using the given temperatures, we can find the relationship between the initial and final volumes: \(\frac{V_1}{300} = \frac{V_2}{200}\) Solving for \(V_2\), we obtain the relationship: \(V_2 = \frac{2}{3} V_1\) So to change the state of the gas while keeping the pressure constant, we should decrease the volume of the gas to two-thirds of its initial value.

Part c: Change in internal energy (∆E)

The internal energy of an ideal gas depends only on its temperature. For an ideal gas, the change in internal energy (∆E) can be expressed as: \(\Delta E = nC_V \Delta T\) Here, \(C_V\) is the heat capacity at constant volume and ∆T is the change in temperature. For the given problem, the temperature falls from 300K to 200K, resulting in: \(\Delta T = T_2 - T_1 = 200 - 300 = -100 \mathrm{~K}\) As the change in internal energy depends solely on temperature, it does not depend on the particular pathway taken to carry out the change of state. No matter if we keep the volume constant or the pressure constant, the change in internal energy will remain the same.