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A system expands from a volume of 1.00 l to 2.00 l against a constant external pressure of 1.00 atm. what is the work (w) done by the system? (1 l·atm = 101.3 j)

Respuesta :

skyluke89
The work done by a system kept at constant pressure is given by:
[tex]W=p \Delta V = p (V_f - V_i)[/tex]
where
p is the pressure
[tex]V_f[/tex] is the final volume
[tex]V_i[/tex] is the initial volume

If we plug the numbers given by the problem into this equation, we find
[tex]W=p (V_f -V_i)=(1.00 atm)(2.00 L-1.00 L)=1.00 L\cdot atm[/tex]

And since [tex]1 atm \cdot L = 101.3 J[/tex], we have that the work done is
[tex]W= 1.00 atm \cdot L = 101.3 J[/tex]
johanrusli

The work done by the system is about 101.3 J

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Further explanation

Let's recall Work Done by Ideal Gas formula as follows:

[tex]\boxed{ W = \int {P} \, dV }[/tex]

where:

P = Gas Pressure ( Pa )

V = Gas Volume ( m³ )

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

initial volume of system = Vo = 1.00 L

final volume of system = V = 2.00 L

constant external pressure = P = 1.00 atm

Asked:

work done by the system = W = ?

Solution:

[tex]W = \int {P} \, dV[/tex]

[tex]W = P \Delta V[/tex] ← Constant External Pressure

[tex]W = P ( V - V_o )[/tex]

[tex]W = 1.00 ( 2.00 - 1.00 )[/tex]

[tex]W = 1.00 ( 1.00 )[/tex]

[tex]W = 1.00 \texttt{ L.atm}[/tex]

[tex]W = 1.00 \times 101.3 \texttt{ J}[/tex] ← Convert from L.atm to Joule

[tex]\boxed {W = 101.3 \texttt{ J} }[/tex]

[tex]\texttt{ }[/tex]

Conclusion:

The work done by the system is about 101.3 J

[tex]\texttt{ }[/tex]

Learn more

  • Buoyant Force : https://brainly.com/question/13922022
  • Kinetic Energy : https://brainly.com/question/692781
  • Volume of Gas : https://brainly.com/question/12893622
  • Impulse : https://brainly.com/question/12855855
  • Gravity : https://brainly.com/question/1724648

[tex]\texttt{ }[/tex]

Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

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