A polytropic process is carried out from an initial pressure of 110 kPa and volume of 5 m³ to a final volume of 2.5 m3. The polytropic index is given by 𝑛 = 1.2. The absolute value of the work done during the process is_______ kJ (round off to 2 decimal places).

The correct answer is 408.92.

Step-by-Step Explanation:

1. Identify the given parameters:
Initial pressure, $P_1=110\text{kPa}$
Initial volume, $V_1=5{\text{m}}^3$
Final volume, $V_2=2.5{\text{m}}^3$
Polytropic index, $n=1.2$

2. Calculate the final pressure ($P_2$):
For a polytropic process, the relationship between pressure and volume is given by:
$P_1{V}_1^n=P_2{V}_2^n$

Substitute the given values into the equation to find $P_2$:
$110\times 5^{1.2}=P_2\times {2.5}^{1.2}$

Rearranging the equation to solve for $P_2$:
$P_2=110\times {\left(\frac{5}{2.5}\right)}^{1.2}=110\times 2^{1.2}$

Using the value $2^{1.2}\approx 2.297397$:
$P_2=110\times 2.297397\approx 252.7136\text{kPa}$

3. Calculate the work done during the process ($W$):
The work done in a polytropic process is calculated using the formula:
$W=\frac{P_1{V}_1-P_2{V}_2}{n-1}$

Substituting the values into this formula:
$W=\frac{110\times 5-252.7136\times 2.5}{1.2-1}$

Calculating the numerator and denominator:
$110\times 5=550$
$252.7136\times 2.5=631.784$
$W=\frac{550-631.784}{0.2}$

Simplifying the fraction:
$W=\frac{-81.784}{0.2}=-408.92\text{kJ}$

4. State the final absolute value:
The absolute value (magnitude) of the work done is:
$\left|W\right|=408.92\text{kJ}$