When 1,2 kg of this gas is heated from 15°C to 210°C at constant pressure, the heat required is 160 kJ. If the characteristic gas constant is R = 0,242 kJ/kg.K, determine the:

Here are the solutions to the problems:

10. Determine the value of the index of expansion $n$.

Step 1: Write down the given values and the polytropic process equation.
The process follows the law $P_1 V_1^n = P_2 V_2^n$.
Given:
$P_1 = 480 \text{ kPa}$
$V_1 = 4.4 \text{ m}^3$
$P_2 = 156.88 \text{ kPa}$
$V_2 = 10.4 \text{ m}^3$

Step 2: Rearrange the equation to solve for $n$.

$$\frac{P_1}{P_2} = \left(\frac{V_2}{V_1}\right)^n$$

Taking the natural logarithm of both sides:

$$\ln\left(\frac{P_1}{P_2}\right) = n \ln\left(\frac{V_2}{V_1}\right)$$ $$n = \frac{\ln\left(\frac{P_1}{P_2}\right)}{\ln\left(\frac{V_2}{V_1}\right)}$$

Step 3: Substitute the given values and calculate $n$.

$$n = \frac{\ln\left(\frac{480 \text{ kPa}}{156.88 \text{ kPa}}\right)}{\ln\left(\frac{10.4 \text{ m}^3}{4.4 \text{ m}^3}\right)}$$ $$n = \frac{\ln(3.060938)}{\ln(2.363636)}$$ $$n = \frac{1.1189}{0.8603}$$ $$n \approx 1.30$$

The value of the index of expansion is $\boxed{1.30}$.

11. Calculate the value of the characteristic gas constant $R$.

Step 1: Convert temperature to Kelvin and pressure to Pascals.
Given:
$P = 120 \text{ kPa} = 120 \times 10^3 \text{ Pa}$
$T = 20^\circ \text{C} = 20 + 273.15 = 293.15 \text{ K}$
$\rho = 1.68 \text{ kg/m}^3$

Step 2: Use the ideal gas law in terms of density to find $R$.
The ideal gas law is $P = \rho R T$.
Rearrange to solve for $R$:

$$R = \frac{P}{\rho T}$$

Step 3: Substitute the values and calculate $R$.

$$R = \frac{120 \times 10^3 \text{ Pa}}{(1.68 \text{ kg/m}^3)(293.15 \text{ K})}$$ $$R = \frac{120000}{492.492}$$ $$R \approx 243.66 \text{ J/(kg} \cdot \text{K)}$$

Converting to $\text{kJ/(kg} \cdot \text{K)}$:

$$R \approx 0.24366 \text{ kJ/(kg} \cdot \text{K)}$$

The characteristic gas constant is $\boxed{0.244 \text{ kJ/(kg} \cdot \text{K)}}$.

12. Determine the specific heat capacities.

Given:
$m = 1.2 \text{ kg}$
$T_1 = 15^\circ \text{C} = 15 + 273.15 = 288.15 \text{ K}$
$T_2 = 210^\circ \text{C} = 210 + 273.15 = 483.15 \text{ K}$
$Q = 160 \text{ kJ}$ (at constant pressure)
$R = 0.242 \text{ kJ/(kg} \cdot \text{K)}$