Give step-by-step solution with explanation and final answer:
3
.

(
2
0
pts
)
Oxygen gas is compressed in a pistoncylinder device from an initial state of
0
.
8
m
3
/
kg and
2
5
\
deg C to a

Question

Give step-by-step solution with explanation and final answer:
3
.
 
(
2
0
 ﻿pts
)
 ﻿Oxygen gas is compressed in a pistoncylinder device from an initial state of 
0
.
8
 ﻿m
3
/
kg and 
2
5
\
deg C to a final state of 
0
.
1
 ﻿m
3
/
kg and 
2
8
7
\
deg C
.
 ﻿Determine the entropy change of the oxygen during this process using the a
)
 ﻿approximate 
(
constant specific heats used
)
,
 ﻿and b
)
 ﻿exact methods 
(
entropy values from Tables used
)
.
 ﻿Discuss any differences. 
(
1
0
 ﻿points
)

VIPSolutions · Accepted Answer

# Entropy Change of Oxygen During Compression

**Given:**
- Initial specific volume, $ v_1 = .8 $ m³/kg  
- Initial temperature, $ T_1 = 25^\circ \text{C} = 298.15 $ K  
- Final specific volume, $ v_2 = .1 $ m³/kg  
- Final temperature, $ T_2 = 287^\circ \text{C} = 560.15 $ K  
- Substance: Oxygen ($ O_2 $), assumed to behave as an ideal gas.

We are to find the entropy change ($ \Delta s $) of oxygen:

---

## a) Approximate Method (Constant Specific Heats)

### Step 1: Use Entropy Change Equation for Ideal Gases

The entropy change per unit mass for an ideal gas, assuming constant specific heats:

$$
\Delta s = c_v \ln \left( \frac{T_2}{T_1} \right) + R \ln \left( \frac{v_2}{v_1} \right)
$$

Alternatively, with $ c_p $ and pressure, but here we use specific volume.

- For $ O_2 $:
  - $ R = .2598 $ kJ/kg·K
  - $ c_v \approx .658 $ kJ/kg·K at room temperature

### Step 2: Substitute Known Values

Plug in the numbers:

- $ T_1 = 298.15 $ K, $ T_2 = 560.15 $ K
- $ v_1 = .8 $ m³/kg, $ v_2 = .1 $ m³/kg
- $ c_v = .658 $ kJ/kg·K
- $ R = .2598 $ kJ/kg·K

$$
\Delta s = .658 \cdot \ln\left(\frac{560.15}{298.15}\right) + .2598 \cdot \ln\left(\frac{.1}{.8}\right)
$$

### Step 3: Calculate Each Term

- $ \ln\left(\frac{560.15}{298.15}\right) = \ln(1.879) = .630 $
- $ \ln\left(\frac{.1}{.8}\right) = \ln(.125) = -2.079 $

Now,

$$
\Delta s = .658 \times .630 + .2598 \times (-2.079)
$$
$$
\Delta s = .414 + (-.540) = -.126 \text{ kJ/kg·K}
$$

---

## b) Exact Method (Using Thermodynamic Tables)

### Step 1: Find Entropy Values from Table A-22 (for $ O_2 $)

Look up in the thermodynamic property tables:

- $ s^\circ_1 = s^\circ(T_1) $ at 298.15 K
- $ s^\circ_2 = s^\circ(T_2) $ at 560.15 K

From standard tables (e.g., Cengel & Boles):

- $ s^\circ(T_1 = 298.15\,K) \approx 6.860 $ kJ/kg·K
- $ s^\circ(T_2 = 560.15\,K) \approx 8.191 $ kJ/kg·K

### Step 2: Apply Entropy Change Formula

For an ideal gas,
$$
\Delta s = \left[ s^\circ(T_2) - s^\circ(T_1) \right] - R \ln\left(\frac{P_2}{P_1}\right)
$$
But we have specific volumes, so use:
$$
\Delta s = \left[ s^\circ(T_2) - s^\circ(T_1) \right] - R \ln\left(\frac{v_2}{v_1}\right)
$$

### Step 3: Substitute Values

$$
\Delta s = (8.191 - 6.860) - .2598 \cdot \ln\left(\frac{.1}{.8}\right)
$$
$$
\ln(.125) = -2.079
$$
$$
\Delta s = 1.331 - .2598 \cdot (-2.079)
$$
$$
\Delta s = 1.331 + .540 = 1.871 \text{ kJ/kg·K}
$$

---

## c) Discussion of Differences

- **Approximate Method:** $\Delta s = -.126$ kJ/kg·K (entropy decreased)
- **Exact Method:** $\Delta s = 1.871$ kJ/kg·K (entropy increased)

**Reason for difference:**  
The approximate method assumes constant specific heats, which is less accurate for large temperature differences, especially for diatomic gases like $ O_2 $ where $ c_v $ increases with temperature. The exact method, based on tabulated entropy values, accounts for this variation, giving a more accurate (and positive) entropy change.

---

## **Final Answers**

**a) Approximate (constant specific heats):**  
$$
\boxed{\Delta s = -.126 \text{ kJ/kg·K}}
$$

**b) Exact (tables):**  
$$
\boxed{\Delta s = 1.87 \text{ kJ/kg·K}}
$$

**c) Discussion:**  
The approximate method underestimates the entropy change and even gives the wrong sign due to the assumption of constant specific heats over a large temperature range. The exact method, using tabulated values, is more reliable for accurate thermodynamic analysis.

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