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Gas in a container is at a pressure of 1.6 * 105 Pa and a volume of 4 m3. What is the work done by the gas if it expands at a constant pressure to three times its initial volume?

Respuesta :

dotmanjohn

Answer:

12.8 Joule

Explanation:

Work done by gas at constant pressure is given by

W = P(ΔV)

W = work done = ?

P = pressure = [tex]1.6X10^{5}[/tex] Pa

V = Volume

(ΔV) = Change in volume

First we calculate for ΔV which is given by the difference between the final and initial volume

(ΔV) = [tex]V_{2} -V_{1}[/tex]

[tex]V_{1}[/tex] = 4 [tex]m^{3}[/tex]

[tex]V_{2}[/tex] = Gas expanded to three times initial volume

Therefore, [tex]V_{2}[/tex] = 3[tex]V_{1}[/tex]

[tex]V_{2}[/tex] = 3 x 4 = 12 [tex]m^{3}[/tex]

(ΔV) = [tex]12 - 4 = 8 m^{3}[/tex]

W = [tex]1.6X10^{5} X 8 = 12.8[/tex] J

Lanuel

The work done by the gas if it expands at a constant pressure to three (3) times its initial volume is [tex]1.28 \times 10^6\;Joules[/tex]

Given the following data:

  • Pressure = [tex]1.6 \times 10^5 \;Pa[/tex]
  • Initial volume = 4 [tex]m^3[/tex]
  • Final volume = [tex]3V_1=3\times 4=12\;m^3[/tex]

To determine the work done by the gas if it expands at a constant pressure to three (3) times its initial volume:

Mathematically, the work done by a gas is given by the formula:

[tex]Work = P \delta V[/tex]

Where:

  • P is the pressure.
  • [tex]\delta V[/tex] is the change in volume.

Substituting the given parameters into the formula, we have;

[tex]Work\;done =1.6 \times 10^5 (12-4)\\\\Work\;done = 1.6 \times 10^5 (8)\\\\Work\;done =1.28 \times 10^6\;Joules[/tex]

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