What is the process from B to C called? Calculate the total work done by an ideal gas in the process ABC.

Here are the solutions to Question 7.3 and 7.4:

QUESTION 7: HEAT

7.3 The volume of a gas at $270 \text{ K}$ increases by $0.29 m^3$ if the temperature increases by $70 \text{ K}$. Calculate the original volume if the pressure remains constant.

Given:

• Initial temperature, $T_1 = 270 \text{ K}$
• Increase in volume, $\Delta V = 0.29 m^3$
• Increase in temperature, $\Delta T = 70 \text{ K}$
• 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 ($V_1$). $T_2 = T_1 + \Delta T = 270 K + 70 K = 340 K$ $V_2 = V_1 + \Delta V = V_1 + 0.29 m^3$

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: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

Step 3: Substitute the known values and solve for the original volume ($V_1$). $\frac{V_1}{270 K} = \frac{V_1 + 0.29 m^3}{340 K}$ Cross-multiply: $340 V_1 = 270 (V_1 + 0.29)$ $340 V_1 = 270 V_1 + 270 \times 0.29$ $340 V_1 = 270 V_1 + 78.3$ $340 V_1 - 270 V_1 = 78.3$ $70 V_1 = 78.3$ $V_1 = \frac{78.3}{70}$ $V_1 \approx 1.1186 m^3$

The original volume is $\boxed{1.12 m^3}$.

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: $P_A = 200 kPa = 200 \times 10^3 \text{ Pa}$, $V_A = 20 m^3$
• Point B: $P_B = 100 kPa = 100 \times 10^3 \text{ Pa}$, $V_B = 50 m^3$ $W_{AB} = Area of trapezoid = \frac{1}{2}(P_A + P_B)(V_B - V_A)$ $W_{AB} = \frac{1}{2}(200 \times 10^3 Pa + 100 \times 10^3 Pa)(50 m^3 - 20 m^3)$ $W_{AB} = \frac{1}{2}(300 \times 10^3 Pa)(30 m^3)$ $W_{AB} = (150 \times 10^3 Pa)(30 m^3)$ $W_{AB} = 4.5 \times 10^6 J = 4500 kJ$

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. $W_{BC} = P \Delta V = P(V_C - V_B) = P(50 m^3 - 50 m^3) = 0 J$

Step 3: Calculate the total work done ($W_{ABC}$). $W_{ABC} = W_{AB} + W_{BC}$ $W_{ABC} = 4.5 \times 10^6 J + 0 J$ $W_{ABC} = 4.5 \times 10^6 J = 4500 kJ$

The total work done by an ideal gas in the process ABC is $\boxed{4500 kJ}$.