give answer in 2 step with explanation at the end of each step and final answer at the end:1. [Thisisawarm-uptoQuestion3.] Consider a monatomic p

ideal gas. Starting from point Bin the figure, the

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:1. [Thisisawarm-uptoQuestion3.] Consider a monatomic p

ideal gas. Starting from point Bin the figure, the process  Fi+-—3A

from B to D has the gas slowly compressed from a volume EH

Vp = 10.0 liters to Vp = 2.0 liters at a constant pressure B Looper

Pg = Pp, =2.0 atm. The gas is then taken through the A 2 B

process from D to A in which some heat is involved while Dj lsobaric i

the gas is held at a constant volume Vz. The pressure 0 2 4 6 8 10V(@L)

and temperature are allowed to rise until point A where the temperature reaches its

original value (point A in the figure has T, = Tp because A and B belong to the same
isotherm).

(a) For the process from B to D, determine how much heat Qz_,p, is involved in terms of
Pg, Vg, and Vp, and state clearly whether heat goes into or gets out of the gas.
Hence, evaluate Qp_,p for the numbers given in the question.

(b) For the process from D to A, determine how much heat Qp_,, is involved in terms of
Pg, Vg, and Vp, and state clearly whether heat goes into or gets out of the gas.
Hence, evaluate Qp_,, for the numbers given

VIPSolutions · Accepted Answer

Let's solve each part step by step, with explanations after each step and the final answers at the end.

---

## (a) **Heat Involved in B → D (Isobaric Compression)**### **Step 1: Expression Terms of Variables**

For an **isobaric process** (constant pressure):

$$
Q_{B \to D} = nC_P\Delta T
$$

Where, for a monatomic ideal gas, $ C_P = \frac{5}{2}R $.

But first, find $\Delta T$:

From the ideal gas law:
$$
P_B V_B = nRT_B \
P_B V_D = nRT_D
$$
So:
$$
T_B = \frac{P_B V_B}{nR}, \quad T_D = \frac{P_B V_D}{nR}
$$
$$
\Delta T = T_D - T_B = \frac{P_B}{nR}(V_D - V_B)
$$

Substitute into $ Q_{B \to D} $:

$$
Q_{B \to D} = nC_P\Delta T = n \left( \frac{5}{2}R \right) \left( \frac{P_B}{nR}(V_D - V_B) \right)
$$
$$
Q_{B \to D} = \frac{5}{2} P_B (V_D - V_B)
$$

**Explanation:**  
Heat is negative if volume decreases (compression), since $ V_D < V_B $, so heat leaves the gas.

---

### **Step 2: Numerical Evaluation**

Given:
- $ P_B = 2. $ atm $ = 2. \times 101325 = 202650 $ Pa
- $ V_B = 10. $ L $ = .010 $ m³
- $ V_D = 2. $ L $ = .002 $ m³

$$
Q_{B \to D} = \frac{5}{2} \times 202650 \times (.002 - .010)
$$
$$
Q_{B \to D} = \frac{5}{2} \times 202650 \times (-.008)
$$
$$
= 2.5 \times 202650 \times -.008
$$
$$
= 506625 \times -.008
$$
$$
= -4053 \text{ J}
$$

**Explanation:**  
Heat ($Q$) is **negative**, so heat leaves the gas during compression.

---

## (b) **Heat Involved in D → A (Isochoric Heating)**

### **Step 1: Expression in Terms of Variables**

For an **isochoric process** (constant volume):

$$
Q_{D \to A} = nC_V\Delta T
$$
Where $ C_V = \frac{3}{2}R $.

From the ideal gas law:
- At D: $ P_B V_D = nRT_D $
- At A: $ P_A V_B = nRT_A $
But $ V = V_B $ is constant, so at D and A:
- At D: $ P_B V_B = nRT_D $
- At A: $ P_A V_B = nRT_A $

But from the problem, at A and B, $ T_A = T_B $.

So,
$$
T_A = T_B, \quad T_D = \frac{P_B V_D}{nR}
$$
So,
$$
\Delta T = T_A - T_D = \frac{P_B V_B}{nR} - \frac{P_B V_D}{nR} = \frac{P_B}{nR}(V_B - V_D)
$$

Substitute into $ Q_{D \to A} $:

$$
Q_{D \to A} = n \left( \frac{3}{2}R \right) \frac{P_B}{nR}(V_B - V_D)
$$
$$
= \frac{3}{2} P_B (V_B - V_D)
$$

**Explanation:**  
Since $ V_B > V_D $, $ \Delta T >  $, heat is **added** to the gas.

---

### **Step 2: Numerical Evaluation**

$$
Q_{D \to A} = \frac{3}{2} \times 202650 \times (.010 - .002)
$$
$$
= 1.5 \times 202650 \times .008
$$
$$
= 303975 \times .008
$$
$$
= 2431.8 \text{ J}
$$

**Explanation:**  
Heat ($Q$) is **positive**, so heat is added to the gas during isochoric heating.

---

## (c) **Work Done, Total Q, and Change in Internal Energy ($\Delta U$) for BDA Cycle**

### **Step 1: Work Done**

- **B → D:** Isobaric compression:
  $$
  W_{B \to D} = P_B (V_D - V_B)
  $$
- **D → A:** Isochoric (constant volume), so $W_{D \to A} = $
- **Total Work:**
  $$
  W_{BDA} = P_B (V_D - V_B)
  $$

$$
W_{BDA} = 202650 \times (.002 - .010) = 202650 \times (-.008) = -1621.2 \text{ J}
$$

**Explanation:**  
Work is **negative** (work done ON the gas, not BY the gas).

---

### **Step 2: Total Q and $\Delta U$ (1st Law)**

Total heat:
$$
Q_{BDA} = Q_{B \to D} + Q_{D \to A} = -4053 + 2431.8 = -1621.2 \text{ J}
$$

Internal energy change ($\Delta U$):
$$
\Delta U = Q + W
$$
$$
\Delta U_{BDA} = -1621.2 + (-1621.2) = -3242.4 \text{ J}
$$

But let's check: Since A and B are on the same isotherm ($T_A = T_B$), the **internal energy depends only on temperature**, so
$$
\Delta U_{B \to A} = 
$$

But numerically, the above calculation adds up Q and W. Since $T_A = T_B$, the correct answer is:
$$
\Delta U_{BDA} = 
$$

**Explanation:**  
For an ideal gas, internal energy only depends on temperature. Since $T_A = T_B$, overall $\Delta U = $.

---

## **Final Answers**

### (a)
$$
Q_{B \to D} = \frac{5}{2} P_B (V_D - V_B)
$$
Numerically:
$$
Q_{B \to D} = -4053 \text{ J} \quad \text{(heat leaves the gas)}
$$

---

### (b)
$$
Q_{D \to A} = \frac{3}{2} P_B (V_B - V_D)
$$
Numerically:
$$
Q_{D \to A} = 2431.8 \text{ J} \quad \text{(heat enters the gas)}
$$

---

### (c)
- **Work done by the gas:** $W_{BDA} = -1621.2 \text{ J}$
- **Total heat involved:** $Q_{BDA} = -1621.2 \text{ J}$
- **Change in internal energy:** $\Delta U_{BDA} = $

---

**Let me know if you want the answers checked with significant digits, or have other questions!**

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