This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here are the solutions to Question 7.3 and 7.4: QUESTION 7: HEAT 7.3 The volume of a gas at 270 K increases by 0.29 m^3 if the temperature increases by 70 K. Calculate the original volume if the pressure remains constant. Given: Initial temperature, T_1 = 270 K Increase in volume, V = 0.29 m^3 Increase in temperature, T = 70 K Pressure remains constant (isobaric process). Step 1: Determine the final temperature (T_2) and express the final volume (V_2) in terms of the original volume (V_1). T_2 = T_1 + T = 270 K + 70 K = 340 K V_2 = V_1 + V = V_1 + 0.29 m^3 Step 2: Apply Charles's Law for an isobaric process. Charles's Law states that for a fixed amount of gas at constant pressure, the volume is directly proportional to its absolute temperature: (V_1)/(T_1) = (V_2)/(T_2) Step 3: Substitute the known values and solve for the original volume (V_1). (V_1)/(270 K) = V_1 + 0.29 m^3340 K Cross-multiply: 340 V_1 = 270 (V_1 + 0.29) 340 V_1 = 270 V_1 + 270 × 0.29 340 V_1 = 270 V_1 + 78.3 340 V_1 - 270 V_1 = 78.3 70 V_1 = 78.3 V_1 = (78.3)/(70) V_1 ≈ 1.1186 m^3 The original volume is 1.12 m^3. 7.4 The pressure-volume diagram of two gas processes is shown in FIGURE 4 below. Study the diagram and answer the questions that follow. 7.4.1 What is the process from B to C called? From the diagram, the volume at point B is 50 m^3 and the volume at point C is also 50 m^3. Since the volume remains constant during the process from B to C, it is an isochoric process. The process from B to C is called an isochoric process (or isovolumetric process). 7.4.2 Calculate the total work done by an ideal gas in the process ABC. The work done by a gas in a P-V diagram is the area under the process curve. Step 1: Calculate the work done during process A to B (W_AB). Process A to B is a linear change in pressure and volume. The work done is the area of the trapezoid under the line AB. Point A: P_A = 200 kPa = 200 × 10^3 Pa, V_A = 20 m^3 Point B: P_B = 100 kPa = 100 × 10^3 Pa, V_B = 50 m^3 W_AB = Area of trapezoid = (1)/(2)(P_A + P_B)(V_B - V_A) W_AB = (1)/(2)(200 × 10^3 Pa + 100 × 10^3 Pa)(50 m^3 - 20 m^3) W_AB = (1)/(2)(300 × 10^3 Pa)(30 m^3) W_AB = (150 × 10^3 Pa)(30 m^3) W_AB = 4.5 × 10^6 J = 4500 kJ Step 2: Calculate the work done during process B to C (W_BC). Process B to C is an isochoric process (constant volume). In an isochoric process, no work is done by the gas. W_BC = P V = P(V_C - V_B) = P(50 m^3 - 50 m^3) = 0 J Step 3: Calculate the total work done (W_ABC). W_ABC = W_AB + W_BC W_ABC = 4.5 × 10^6 J + 0 J W_ABC = 4.5 × 10^6 J = 4500 kJ The total work done by an ideal gas in the process ABC is 4500 kJ. Got more? Send 'em.