A volume of $$ 0,11 \mathrm{~m}^{s} $$ of a certain gas is initially at a temperature of $$ 170^{\circ} \mathrm{C} $$ and a pressure of 290 kPa . It is then cooled down. at constant pressure, to a temperature of $$ 30^{\circ} \mathrm{C} $$. Calculate what amount of heat is transmitted by the gas and what amount of work is done on the gas during the process. For this gas is $$ \mathrm{R}=0,29 \mathrm{~kJ} / \mathrm{kgk} $$ and Cp $$ =1,005 \mathrm{~kJ} / \mathrm{kg} \mathrm{K} $$.

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To solve this problem, we'll analyze the cooling process of the gas using the principles of thermodynamics, particularly focusing on **heat transfer** and **work done** during a **constant pressure process**.

### **Given Data:**
- **Initial Volume ($$ V_1 $$)**: $$ 0.11 \, \text{m}^3 $$
- **Initial Temperature ($$ T_1 $$)**: $$ 170^\circ\text{C} $$ → $$ 443 \, \text{K} $$
- **Final Temperature ($$ T_2 $$)**: $$ 30^\circ\text{C} $$ → $$ 303 \, \text{K} $$
- **Pressure ($$ P $$)**: $$ 290 \, \text{kPa} $$ (constant)
- **Gas Constant ($$ R $$)**: $$ 0.29 \, \text{kJ/kg·K} $$
- **Specific Heat at Constant Pressure ($$ C_p $$)**: $$ 1.005 \, \text{kJ/kg·K} $$

### **Step 1: Determine the Mass of the Gas ($$ m $$)**

Using the ideal gas law at constant pressure:
$$
PV = mRT
$$
Solving for mass ($$ m $$):
$$
m = \frac{PV}{RT}
$$

Plugging in the initial conditions:
$$
m = \frac{290 \, \text{kPa} \times 0.11 \, \text{m}^3}{0.29 \, \text{kJ/kg·K} \times 443 \, \text{K}} \approx 0.249 \, \text{kg}
$$

### **Step 2: Calculate the Heat Transferred ($$ Q $$)**

For a constant pressure process, the heat transferred is given by:
$$
Q = mC_p\Delta T
$$
Where:
$$
\Delta T = T_2 - T_1 = 303 \, \text{K} - 443 \, \text{K} = -140 \, \text{K}
$$

Thus:
$$
Q = 0.249 \, \text{kg} \times 1.005 \, \text{kJ/kg·K} \times (-140 \, \text{K}) \approx -35.0 \, \text{kJ}
$$
**Interpretation:** A negative sign indicates that **35.0 kJ of heat is released** by the gas.

### **Step 3: Calculate the Work Done on the Gas ($$ W $$)**

Work done during a constant pressure process is:
$$
W = P\Delta V
$$
First, determine the final volume ($$ V_2 $$) using the ideal gas law:
$$
\frac{V_1}{T_1} = \frac{V_2}{T_2} \Rightarrow V_2 = V_1 \times \frac{T_2}{T_1} = 0.11 \, \text{m}^3 \times \frac{303}{443} \approx 0.0752 \, \text{m}^3
$$
$$
\Delta V = V_2 - V_1 = 0.0752 \, \text{m}^3 - 0.11 \, \text{m}^3 = -0.0348 \, \text{m}^3
$$
$$
W = 290 \, \text{kPa} \times (-0.0348 \, \text{m}^3) \approx -10.1 \, \text{kJ}
$$
**Interpretation:** A negative sign for $$ W $$ implies that work is **done on the gas**. Therefore, **10.1 kJ of work is done on the gas**.

### **Final Answers:**
- **Heat Transferred ($$ Q $$)**: $$ -35.0 \, \text{kJ} $$ (35.0 kJ of heat is released by the gas)
- **Work Done on the Gas ($$ W $$)**: $$ +10.1 \, \text{kJ} $$
The gas releases 35.0 kJ of heat and 10.1 kJ of work is done on it during the cooling process.

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