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7. Figure 3 shows the expansion of an ideal gas from initial volume V; = 0.25 m³ and pressure P = 200 kPa to final volume V; = 1 m³ and pressure Pf = 50 kPa. P (kPa) 200 150 100 a) The process depicted in Figure 3 is one of four the following fundamental 50 processes: isobaric, isochoric, isothermal, and adiabatic. Which type of process is it? Why? Explain your answer. V (m² ) 0.25 0.5 0.75 Figure 3 b) Calculate the work done in this process. Show your work.

7. Figure 3 shows the expansion of an ideal gas from initial volume V; = 0.25 m³ and pressure P = 200 kPa to final volume V; = 1 m³ and pressure Pf = 50 kPa. P (kPa) 200 150 100 a) The process depicted in Figure 3 is one of four the following fundamental 50 processes: isobaric, isochoric, isothermal, and adiabatic. Which type of process is it? Why? Explain your answer. V (m² ) 0.25 0.5 0.75 Figure 3 b) Calculate the work done in this process. Show your work.

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### Problem Statement Figure 3 shows the expansion of an ideal gas from an initial volume \( V_i = 0.25 \, m^3 \) and pressure \( P_i = 200 \, kPa \) to a final volume \( V_f = 1 \, m^3 \) and pressure \( P_f = 50 \, kPa \). a) The process depicted in Figure 3 is one of the following four fundamental processes: isobaric, isochoric, isothermal, and adiabatic. Which type of process is it? Why? Explain your answer. b) Calculate the work done in this process. Show your work. ### Diagram Explanation The graph (Figure 3) is a plot of pressure \( P \) (in kilopascals) on the vertical axis versus volume \( V \) (in cubic meters) on the horizontal axis. - The curve starts at a high pressure and low volume point (200 kPa, 0.25 m\(^3\)) and ends at a low pressure and high volume point (50 kPa, 1 m\(^3\)). - The curve slopes downwards, indicating that as the volume increases, the pressure decreases. - This shape is characteristic of an isothermal process, where the temperature remains constant as the ideal gas expands. ### Explanation for Part (a) The process shown in the graph is isothermal. This conclusion is based on the nature of the graph, which shows a hyperbolic relationship between pressure and volume, consistent with the ideal gas law \( PV = nRT \). In an isothermal process, the temperature remains constant, suggesting the product of pressure and volume is constant for a given amount of gas. ### Solution for Part (b) To calculate the work done in an isothermal process, use the formula: \[ W = nRT \ln\left(\frac{V_f}{V_i}\right) \] If the number of moles \( n \) and the temperature \( T \) are not given, we can express the work done as: \[ W = \int_{V_i}^{V_f} P \, dV = nRT \ln\left(\frac{V_f}{V_i}\right) \] Under the assumption that the ideal gas is involved, and using the given pressures and volumes, the work done can be evaluated numerically, typically using specific gas
Transcribed Image Text:### Problem Statement Figure 3 shows the expansion of an ideal gas from an initial volume \( V_i = 0.25 \, m^3 \) and pressure \( P_i = 200 \, kPa \) to a final volume \( V_f = 1 \, m^3 \) and pressure \( P_f = 50 \, kPa \). a) The process depicted in Figure 3 is one of the following four fundamental processes: isobaric, isochoric, isothermal, and adiabatic. Which type of process is it? Why? Explain your answer. b) Calculate the work done in this process. Show your work. ### Diagram Explanation The graph (Figure 3) is a plot of pressure \( P \) (in kilopascals) on the vertical axis versus volume \( V \) (in cubic meters) on the horizontal axis. - The curve starts at a high pressure and low volume point (200 kPa, 0.25 m\(^3\)) and ends at a low pressure and high volume point (50 kPa, 1 m\(^3\)). - The curve slopes downwards, indicating that as the volume increases, the pressure decreases. - This shape is characteristic of an isothermal process, where the temperature remains constant as the ideal gas expands. ### Explanation for Part (a) The process shown in the graph is isothermal. This conclusion is based on the nature of the graph, which shows a hyperbolic relationship between pressure and volume, consistent with the ideal gas law \( PV = nRT \). In an isothermal process, the temperature remains constant, suggesting the product of pressure and volume is constant for a given amount of gas. ### Solution for Part (b) To calculate the work done in an isothermal process, use the formula: \[ W = nRT \ln\left(\frac{V_f}{V_i}\right) \] If the number of moles \( n \) and the temperature \( T \) are not given, we can express the work done as: \[ W = \int_{V_i}^{V_f} P \, dV = nRT \ln\left(\frac{V_f}{V_i}\right) \] Under the assumption that the ideal gas is involved, and using the given pressures and volumes, the work done can be evaluated numerically, typically using specific gas
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