Two Ways from A to B

solved FirstLawOfThermodynamics

Q. In changing the state of a gas adiabatically $Δ Q (a) = 0$ J from an equilibrium state A to another equilibrium state B $Δ U (t) = Δ U (a)$ , an amount of work equal to 22.3 J $W_{on} (a) = 22.3$ J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 39.18 J $Δ Q (t) = 39.18$ J, how much is the net work done by the system in the latter case, in joules?

Inputs

work on

W_{on}

22.3 J

heat

Δ Q

39.18 J

Outputs

work by

Δ W

How it's solved

Solved with FirstLawOfThermodynamics

Given

$Δ Q (t) = 39.18$ J
$W_{on} (a) = 22.3$ J
$Δ Q (a) = 0$ J from “adiabatically”
$Δ U (t) = Δ U (a)$ from “from an equilibrium state A to another equilibrium state B”

Find $Δ W (t)$ .

Working

work on is work by turned round(adiabatically) $W_{on} (a) = - Δ W (a)$
$22.3 = - Δ W (a)$
$Δ W (a) = −22.3$ J
first law(adiabatically) $Δ Q (a) = Δ U (a) + Δ W (a)$
$0 = Δ U (a) + (−22.3)$
$Δ U (a) = 22.3$ J
from an equilibrium state A to another equilibrium state B $Δ U (t) = Δ U (a)$
$Δ U (t) = 22.3$
$Δ U (t) = 22.3$ J
first law(the other way) $Δ Q (t) = Δ U (t) + Δ W (t)$
$39.18 = 22.3 + Δ W (t)$
$Δ W (t) = 16.88$ J
$Δ W (t) \approx 16.9$ Jto 3 significant figures

Conclusion

Internal energy depends only on the state, so both paths from A to B change it by the same amount; the heat the second path takes in, less that change, is the work it does.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :two_ways_from_a_to_b do4  kind :numerical5  describes "In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state " \6            "B, an amount of work equal to %{work_on} is done on the system. If the gas is taken from state A to B " \7            "via a process in which the net heat absorbed by the system is %{heat}, %{work_by}, in joules?"8  given work_on: "22.3 J"9  given heat: "39.18 J"10  asks work_by: "how much is the net work done by the system in the latter case"11end

solution.rb

1require_relative "../questions/two_ways_from_a_to_b"2require_relative "../science/thermodynamics/first_law_of_thermodynamics"34solve :two_ways_from_a_to_b, using: FirstLawOfThermodynamics do5  objects :gas6  gas.goes(:adiabatically)7  gas.goes(:the_other_way)89  take heat: gas.heat_absorbed(during: the_other_way)10  take work_on: gas.work_done_on_the_system(during: adiabatically)1112  deduce_from "adiabatically", gas.heat_absorbed(during: adiabatically) => "0 J"13  deduce_from "from an equilibrium state A to another equilibrium state B",14              gas.change_of_internal_energy(during: the_other_way) => gas.change_of_internal_energy(during: adiabatically)1516  work_by = gas.work_done_by_the_system(during: the_other_way)17  answer work_by:, in: "J", figures: 31819  conclusion "Internal energy depends only on the state, so both paths from A to B change it by the same " \20             "amount; the heat the second path takes in, less that change, is the work it does."21end

playground.rb

1require_relative "../solutions/two_ways_from_a_to_b"23play_with :two_ways_from_a_to_b do4  title "Two Ways from A to B"5end

The laws it drew on

quantity	written	units	within	from
heat absorbed	$Δ Q$	J	—	FirstLawOfThermodynamics
work done by the system	$Δ W$	J	—	FirstLawOfThermodynamics
work done on the system	$W_{on}$	J	—	FirstLawOfThermodynamics
change of internal energy	$Δ U$	J	—	FirstLawOfThermodynamics
heat absorbed of gas during adiabatically	$Δ Q (a)$	J	—
change of internal energy of gas during adiabatically	$Δ U (a)$	J	—
work done by the system of gas during adiabatically	$Δ W (a)$	J	—
heat absorbed of gas during the other way	$Δ Q (t)$	J	—
change of internal energy of gas during the other way	$Δ U (t)$	J	—
work done by the system of gas during the other way	$Δ W (t)$	J	—
work done on the system of gas during adiabatically	$W_{on} (a)$	J	—
work done on the system of gas during the other way	$W_{on} (t)$	J	—

Δ Q (a) = Δ U (a) + Δ W (a)

first law(adiabatically)

Δ Q (t) = Δ U (t) + Δ W (t)

first law(the other way)

W_{on} (a) = - Δ W (a)

work on is work by turned round(adiabatically)

W_{on} (t) = - Δ W (t)

work on is work by turned round(the other way)

Δ U (t) = Δ U (a)

from an equilibrium state A to another equilibrium state B