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4.8 m³ of air at 20 kPa gauge and 20 degrees C is compressed at constant temperature to a final pressure of 320 kPa gauge.

1. What volume will the air occupy at the final pressure? (Take the atmospheric pressure to be 98.3 kPa)

2. What will be the final temperature? (The final temperature in degrees Celsius)

Answer :

Using Boyle's Law, we determined the final volume of the air to be 1.36 m³ when compressed to a final pressure of 418.3 kPa. The temperature remains constant at 20°C.

To solve this problem, we will use the principles of Boyle's Law, which states that for a given mass of gas at constant temperature, the product of pressure and volume is constant (P1V1 = P2V2).

Part 1: Volume at Final Pressure

First, we convert the gauge pressures to absolute pressures by adding atmospheric pressure (98.3 kPa):

  • Initial absolute pressure (P1) = 20 kPa + 98.3 kPa = 118.3 kPa
  • Final absolute pressure (P2) = 320 kPa + 98.3 kPa = 418.3 kPa

Using Boyle's Law:

[tex]\(P1V1 = P2V2\)[/tex]

Where:

  • [tex]\(P1 = 118.3 kPa\)[/tex]
  • [tex]\(V1 = 4.8 m^3\)[/tex]
  • [tex]\(P2 = 418.3 kPa\)[/tex]

[tex]\(118.3\ kPa \times 4.8\ m^3 = 418.3\ kPa \times V2\)[/tex]

Solving for (V2):

[tex]\(V2 =\ \frac{118.3\ kPa \times 4.8\ m^3}{418.3\ kPa} = 1.36\ m^3\)[/tex]

Part 2: Final Temperature

Since the question states that the compression occurs at constant temperature, the final temperature remains the same as the initial temperature:

  • Initial temperature = 20°C
  • Final temperature = 20°C

Therefore, the final temperature is also 20°C.

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Rewritten by : Barada