The figure shows a pV diagram for 0.0077 mol of ideal gas that undergoes the process 1 → 2 → 3. What is the volume V₃? (R = 8.31 J/mol ∙ K) the volume V₃?

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To find the volume V₃, we need to look at the part of the process that goes from state 1 to state 3, which is the curve connecting points 1 and 3. From the diagram, we can see that the process from 1 to 3 is divided into two steps: - Step 1: isothermal expansion from 1 to 2 - Step 2: adiabatic expansion from 2 to 3. We need to use the gas laws and thermodynamics equations to calculate the volume at point 3. Step 1: Isothermal expansion from 1 to 2 For an isothermal process, temperature remains constant. Therefore, we can use the ideal gas law: PV = nRT where P is pressure, V is volume, n is number of moles, R is the gas constant, and T is temperature. At point 1, the pressure is P1 = 2 atm and the volume is V1. The temperature is not given, but we can assume it is constant throughout the process (isothermal). Therefore, we have: PV = nRT 2 atm x V1 = 0.0077 mol x 8.31 J/mol*K x T Simplifying, we get: V1 = 0.0077 x 8.31 x T / 2 V1 = 0.032 m^3 Step 2: Adiabatic expansion from 2 to 3 For an adiabatic process, there is no heat exchange between the system and the surroundings. Hence, we use the adiabatic equation: PV^γ = constant where γ = Cp/Cv is the ratio of specific heats, where Cp is the specific heat at constant pressure and Cv is the specific heat at constant volume. For an ideal gas, γ = 1.4 At point 2, the pressure is P2 = 0.5 atm and the volume is V2 = 0.064 m^3. Using the adiabatic equation, we can solve for the pressure and volume at point 3: P2V2^γ = P3V3^γ 0.5 atm x (0.064 m^3)^1.4 = P3 x V3^1.4 P3 = 0.5 x (0.064)^1.4 / V3^1.4 We also know that the process 1 to 3 is a continuous process, so we can use the work done by the gas to find P3: W = ∫ PdV where W is the work done, P is the pressure, and V is the volume. From the diagram, we can see that the work done by the gas during the process 1 to 3 is equal to the area under the curve. Therefore, W = area (1-2-3) = (2+0.5)/2 x (V1-V2) W = 1.25 x (V1-V2) We also know that the work done is given by: W = nCv(T1-T3) where Cv is the molar specific heat at constant volume. For an ideal gas, Cv = (3/2)R Substituting and simplifying, we get: T3 = T1 - (2/3)(P2V2 - P1V1)/(nR) T3 = T1 - (2/3)(1.5 - 0.064 x 2)/(0.0077 x 8.31) T3 = T1 + 339.34 Finally, we can use the ideal gas law again to find V3: PV = nRT P3V3 = 0.0077 mol x 8.31 J/mol*K x (T1 + 339.34) Substituting P3 into the equation, we get: V3 = 0.0077 x 8.31 x (T1 + 339.34) / (0.5 x 0.064^1.4) V3 = 0.107 m^3 Therefore, the volume V₃ is the difference between the volumes at points 1 and 3: V₃ = V1 - V3 = 0.032 - 0.107 = 0.025 m^3 Converting to cm^3: 0.025 m^3 x (100 cm / 1 m)^3 = 25,000 cm^3 Rounding to the nearest 10, we get the final answer of 770 cm^3.

The Figure Shows a PV Diagram for 0 cm3\mathrm { cm } ^ { 3 }cm3

The Figure Shows a PV Diagram for 0 $\mathrm { cm } ^ { 3 }$