Calculate the work energy, w, gained or lost by the system when a gas expands from 15 l to 35 l against a constant external pres

Question

MrMuchimi

4 years ago

13

Calculate the work energy, w, gained or lost by the system when a gas expands from 15 l to 35 l against a constant external pres

sure of 1.5 atm

Physics

2 answers:

Irina-Kira [14] · Answer 1 · 2021-11-19T03:44:32+03:00

The work done by the gas is about 30 L.atm

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<h3>Further explanation</h3>

The Ideal Gas Law and Work formula that needs to be recalled is:

$\boxed {PV = nRT}$

$\boxed { W = P \Delta V }$

<em>where:</em>

<em>P = Pressure ( Pa )</em>

<em>V = Volume ( m³ )</em>

<em>n = number of moles ( moles )</em>

<em>R = Gas Constant ( 8.314 J/mol K )</em>

<em>T = Absolute Temperature ( K )</em>

Let us now tackle the problem !

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<u>Given:</u>

initial volume = Vo = 15 L

final volume = V = 35 L

constant pressure = P = 1.5 atm

<u>Asked:</u>

word done by the gas = W = ?

<u>Solution:</u>

$W = P \Delta V$

$W = P \times ( V - Vo )$

$W = 1.5 \times ( 35 - 15 )$

$W = 1.5 \times 20$

$W = 30 \texttt{ L.atm }$

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<h3>Conclusion:</h3>

The work done by the gas is about 30 L.atm

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<h3>Learn more</h3>

Minimum Coefficient of Static Friction : brainly.com/question/5884009
The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

AnnyKZ [126] · Answer 2 · 2021-11-19T02:38:23+03:00

In physical chemistry or in thermodynamics, the work done on the system or by the system (depending on the sign convention) can be determined in several ways. When assumptions like ideal gas behavior is applied, then the formula for work is

W = Δ(PV)

which is the change of the product of Pressure and Volume. But since it was specified that Pressure is constant, the work could be simplified into

W = PΔV = P(V₂ - V₁)

Since we already know the constant pressure and the volumes of the ideal gas before and after the change, we could now solve for work. But let's establish first the units of work which is in Joules. When simplified, Joules is equal to m³*Pa. So, we first change the unit of pressure from atm to Pascals ( 1 atm = 101,325 Pa) and the unit of volume from liters to m³ (1 m³ = 1000 L),

1.5 atm * 101325 Pa/1 atm = 151987.5 Pa
15 L * 1 m³/1000 L = 0.015 m³
35 L * 1 m³/1000 L = 0.035 m³

Then, they are now ready for substitution,

W = 151987.5 Pa (0.035 m³ - 0.015 m³)
W = 3,039.75 Joules