Answers
The work done by the gas is about 30 L.atm
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Further explanationThe Ideal Gas Law and Work formula that needs to be recalled is:
[tex]\boxed {PV = nRT}[/tex]
[tex]\boxed { W = P \Delta V }[/tex]
where:
W = Work ( J )
P = Pressure ( Pa )
V = Volume ( m³ )
n = number of moles ( moles )
R = Gas Constant ( 8.314 J/mol K )
T = Absolute Temperature ( K )
Let us now tackle the problem !
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Given:
initial volume = Vo = 15 L
final volume = V = 35 L
constant pressure = P = 1.5 atm
Asked:
word done by the gas = W = ?
Solution:
[tex]W = P \Delta V[/tex]
[tex]W = P \times ( V - Vo )[/tex]
[tex]W = 1.5 \times ( 35 - 15 )[/tex]
[tex]W = 1.5 \times 20[/tex]
[tex]W = 30 \texttt{ L.atm }[/tex]
[tex]\texttt{ }[/tex]
Conclusion:The work done by the gas is about 30 L.atm
[tex]\texttt{ }[/tex]
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Answer detailsGrade: High School
Subject: Physics
Chapter: Pressure
The work energy, w, gained or lost by the system when the gas expands is 30 Joules.
Given the following data:
Pressure = 1.5 atmInitial volume = 15 litersFinal volume = 35 litersTo find the work energy, w, gained or lost by the system when the gas expands:
Mathematically, the work energy by a system is given by the formula:
[tex]Work = P \delta V[/tex]
Where:
P is the pressure.V is the volume.Substituting the given parameters into the formula, we have;
[tex]Work\;energy = 1.5(35-15)\\\\Work\;energy = 1.5(20)[/tex]
Work energy = 30 Joules.
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Work is the product of force with the displacement of objects.
Can be formulated
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work done by the applied force = Wp work done by the frictional force = Wfthe magnitude is the same as the force applied by the applied force (Fp) but the opposite direction
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In this circumstance, the sand core is completely submerged and the fluid (molten steel) that fills its cavity takes its form after solidifying. The given buoyancy force (225 N) is essentially the weight of the fluid displaced by it ( weight = mass x gravity). The weight of the mold cavity would hence be the weight of the steel casting.
Given the volume of the mold cavity (4840 cm
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The mass (m) would hence be: m = volume x density = 7.85 g/cm3 x 4840 cm3 = 37974 g = 37.974 kg.
Therefore, the weight of the final steel casting is approximately 37.974 kg.
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