Chemistry, 13.02.2020 22:22 haileyrae187

One mole of an ideal gas is contained in a cylinder with a movable piston. The temperature is constant at 778C. Weights are removed suddenly from the piston to give the following sequence of three pressures: a. P1 5 5.00 atm (initial state) b. P2 5 2.24 atm c. P3 5 1.00 atm (final state)

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Answer from: kaiya2022547

One mole of an ideal gas is contained in a cylinder with a movable piston. The temperature is constant at 77 °C. Weights are removed suddenly from the piston to give the following sequence of three pressures.

a. P1 = 5.50 atm (initial state)

b. P2 = 2.43 atm

c. P3 = 1.00 atm (final state)

What is the total work (in joules) in going from the initial to the final state by way of the preceding two steps?

The answer to the question is

The total work (in joules) in going from the initial to the final state by way of the preceding two steps is equal to -4685.6 Joules

Explanation:

For a constant temperature or isothermal process, we have

P₁V₁ = P₂V₂ = P₃V₃ which gives $\frac{V_{1} }{V_{3} } = \frac{P_{3} }{P_{1} }$

From the first law of thermodynamics, Q =ΔU+W and ΔU = 0 therefore

Q = W and the work done is given by Q = n·R·T· $ln(\frac{V_{1} }{V_{3} } )$ = $nRTln(\frac{P_{3} }{P_{1} } )$

Where

T = 77 °C = 350.15 K

P₁ = 5.00 atm

P₂ = 2.24 atm

P₃ = 1.00 atm

n = 1 mole

R = 8.3145 J/mol·K

We are required to find the the total work (in joules) in going from the initial to the final state by way of the preceding two steps

Therefore we have

Total work done is given by

$W = -(nRT ln(\frac{P_{1} }{P_{2} })+nRTln(\frac{P_{2} }{P_{3} })$ or $= -nRT( ln(\frac{P_{1} }{P_{2} })+ln(\frac{P_{2} }{P_{3} })$

Which by the logarithm laws ㏑(A)+㏑(B) = ㏑(A×B) the equation becomes

$W= -nRT( ln(\frac{P_{1} }{P_{2} }*\frac{P_{2} }{P_{3} }))$ = $-nRT( ln(\frac{P_{1} }{P_{3} }}))$ therefore the equation is the same for calculating directly from the initial pressure P₁, to the final pressure P₃

Therefore W = (1.00 mol)×(8.3145 J/mol·K)× $ln(\frac{5 atm}{1 atm} )$

= -4685.6 J or -4.69 kJ

The total work (in joules) in going from the initial to the final state by way of the preceding two steps is the same as the work arrived at by calculating directly from the initial to the final state bypassing the intermediate state, P₂ and is equal to -4685.6 Joules. This is so because P₁V₁ = P₂V₂ = P₃V₃

Answer from: Quest

the answer is true. hope it .

Answer from: Quest

it should have an internal chamber and two thermometers sticking out of the top. the internal chamber will also be closed off from the outside.

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