3. 3m^3 of air expands from 0.54 MPa and 45^circ C to 0.15 MP a in a polytropic process with an exponent of n=1.36 Find the final volume and temperature of the air.

To solve this problem, we can use the polytropic process equation:<br /><br />\[ P_1 V_1^n = P_2 V_2^n \]<br /><br />where:<br />- \( P_1 \) is the initial pressure (0.54 MPa)<br />- \( V_1 \) is the initial volume (3 m³)<br />- \( P_2 \) is the final pressure (0.15 MPa)<br />- \( V_2 \) is the final volume<br />- \( n \) is the polytropic exponent (1.36)<br /><br />First, we need to find the final volume \( V_2 \):<br /><br />\[ V_2 = \frac{P_1 V_1^n}{P_2} \]<br /><br />Substituting the given values:<br /><br />\[ V_2 = \frac{(0.54 \text{ MPa}) \times (3 \text{ m}^3)^{1.36}}{0.15 \text{ MPa}} \]<br /><br />\[ V_2 = \frac{0.54 \times 3^{1.36}}{0.15} \]<br /><br />\[ V_2 = \frac{0.54 \times 4.080}{0.15} \]<br /><br />\[ V_2 = \frac{2.2032}{0.15} \]<br /><br />\[ V_2 = 14.688 \text{ m}^3 \]<br /><br />Next, we need to find the final temperature \( T_2 \). We can use the relation for a polytropic process:<br /><br />\[ \frac{T_2}{T_1} = \left( \frac{V_2}{V_1} \right)^{n-1} \]<br /><br />First, we need to convert the initial temperature \( T_1 \) from Celsius to Kelvin:<br /><br />\[ T_1 = 45^{\circ}C + 273.15 = 318.15 \text{ K} \]<br /><br />Now, we can find \( T_2 \):<br /><br />\[ T_2 = T_1 \times \left( \frac{V_2}{V_1} \right)^{n-1} \]<br /><br />\[ T_2 = 318.15 \times \left( \frac{14.688}{3} \right)^{1.36-1} \]<br /><br />\[ T_2 = 318.15 \times \left( 4.896 \right)^{0.36} \]<br /><br />\[ T_2 = 318.15 \times 1.898 \]<br /><br />\[ T_2 = 603.57 \text} \, the final volume of the air is 14.688 m³, and the final temperature is 603.57 K.