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 b
y the system? (1 l·atm = 101.3 j)
2 answers:
The amount of work performed by a system at consistent pressure is defined by the following equation:
where
p represents pressure
as the final volume
as the initial volume
Plugging the values given in this case into the formula gives us
Considering that
, the result for the work done becomes
where
p represents pressure
Plugging the values given in this case into the formula gives us
Considering that
The system has done work of approximately 101.3 J
Additional explanation
To revisit Work Done by Ideal Gas formula can be expressed as:
where:
P = Gas Pressure ( Pa )
V = Gas Volume ( m³ )
Next, let us look at the problem!
Given:
initial volume of system = Vo = 1.00 L
final volume of system = V = 2.00 L
constant external pressure = P = 1.00 atm
What we need to find is:
work done by the system = W =?
Solution:
← Constant External Pressure
←
to Joule
Conclusion:
The system's work amounts to approximately 101.3 J
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Answer details
Grade: High School
Subject: Physics
Chapter: Pressure
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Response:
The primary consequence is an increase in induced charge at the nearest points. However, the overall net charge remains zero, meaning it does not influence the flow.
We can utilize Gauss's law to solve this problem
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According to Gauss's law, if the sphere is offset, the only effect it generates is an increment in induced charge at the nearest points. Nevertheless, the net charge remains zero, so it does not impact the flow; irrespective of the sphere's position, the total induced charge is consistently equal to the charge on the sphere.
Response:
The work performed by the particle traveling from x = 0 to x = 2 m totals 20 J.
Details:
The force impacting a particle, which is restricted to the x-axis, is expressed as follows:
We need to calculate the work done on a particle moving from x = 0.00 m to x = 2.00 m.
The formula for the work done by the particle is defined as:
Consequently, the work executed by the particle between x = 0 and x = 2 m amounts to 20 J. Thus, this is the solution sought.
Let "L" denote the beam's length.
According to the diagram:
AD = length of the beam = L
AC = CD = AD/2 = L/2
BC = AC - AB = (L/2) - 1.10
BD = AD - AB = L - 1.10
m = mass of the beam = 20 kg
m₁ = mass of the child at one end = 30 kg
m₂ = mass of the child at the opposite end = 40 kg
By applying the principle of torque equilibrium around point B:
(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)
30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)
L = 1.98 m
