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Консультация № 193785
30.10.2018, 10:15
0.00 руб.
01.11.2018, 09:20
1
5
1
Уважаемые эксперты! Пожалуйста, проверьте решение.
На участках 1-2 и 3-4 работа не совершается, а изменение внутренней энергии равны и противоположны по знаку, значит в сумме дают 0.
Тогда газ получал тепло только на участке 2-3.
Q=[$916$]U+A
A=P2(V2-V1)=2*10^6*10^-3=2кДж
P2V1=VRT2
P2V2=VRT3
P2(V2-V1)=VR[$916$]T
[$916$]U=3/2VR[$916$]T=P2(V2-V1)=3/2*2*10^6*10^-3=3кДж
Q=5кДж
На участках 1-2 и 3-4 работа не совершается, а изменение внутренней энергии равны и противоположны по знаку, значит в сумме дают 0.
Тогда газ получал тепло только на участке 2-3.
Q=[$916$]U+A
A=P2(V2-V1)=2*10^6*10^-3=2кДж
P2V1=VRT2
P2V2=VRT3
P2(V2-V1)=VR[$916$]T
[$916$]U=3/2VR[$916$]T=P2(V2-V1)=3/2*2*10^6*10^-3=3кДж
Q=5кДж
Прикрепленные файлы:
Обсуждение
30.10.2018, 14:52
общий
30.10.2018, 14:53
Изменения внутренней энергии на участках 1-2 и 3-4 не будут равны по модулю, поскольку температура прямо пропорциональна как давлению, так и объёму.
Предложу следующий подход:
Внутренняя энергия - термодинамическая величина и не зависит от пути процесса, поэтому можете сразу найти её изменение в процессе 1-4.
А с работой Вы верно определили, что она совершается только на участке 2-3.
Предложу следующий подход:
Внутренняя энергия - термодинамическая величина и не зависит от пути процесса, поэтому можете сразу найти её изменение в процессе 1-4.
А с работой Вы верно определили, что она совершается только на участке 2-3.
30.10.2018, 18:04
общий
Адресаты:
Тогда
1-2: V1(P2-P1)=VR∆T12
∆U=3/2*V1(P2-P1)=1650Дж
3-4:. V2(P1-P2)=VP∆T34
∆U=3/2*V2(P1-P2)=-3300Дж
Q=5000+1650-3300=3350Дж
1-2: V1(P2-P1)=VR∆T12
∆U=3/2*V1(P2-P1)=1650Дж
3-4:. V2(P1-P2)=VP∆T34
∆U=3/2*V2(P1-P2)=-3300Дж
Q=5000+1650-3300=3350Дж
31.10.2018, 01:05
общий
Адресаты:
Да, это верный ответ.
Альтернативный путь решения, как я указывал:
[$916$]U=1.5*(p1V2-p1V1)=1.35 кДж (сразу между начальной и конечной точкой процесса)
A=p2*(V2-V1)=2кДж (работа совершается только на участке 2-3)
Q=[$916$]U+A=3.35 кДж
Альтернативный путь решения, как я указывал:
[$916$]U=1.5*(p1V2-p1V1)=1.35 кДж (сразу между начальной и конечной точкой процесса)
A=p2*(V2-V1)=2кДж (работа совершается только на участке 2-3)
Q=[$916$]U+A=3.35 кДж
01.11.2018, 09:28
общий
это ответ
Здравствуйте, Кожухова Дарья!
Как мы рассмотрели в мини-форуме, изменения внутренней энергии на участках 1-2 и 3-4 не равны по модулю, поскольку температура прямо пропорциональна объёму. Поэтому их также необходимо учесть в расчётах.
С другой стороны, внутренняя энергия является термодинамической величиной, поэтому её изменение не зависит от пути процесса и может быть найдено сразу между начальным и конечным состояниями:
[$916$]U=1.5[$183$](p1V2-p1V1)=1.35 кДж
Как Вы верно определили, работа совершается только на участке 2-3
A=p2[$183$](V2-V1)=2 кДж
Q=[$916$]U+A=3.35 кДж
Как мы рассмотрели в мини-форуме, изменения внутренней энергии на участках 1-2 и 3-4 не равны по модулю, поскольку температура прямо пропорциональна объёму. Поэтому их также необходимо учесть в расчётах.
С другой стороны, внутренняя энергия является термодинамической величиной, поэтому её изменение не зависит от пути процесса и может быть найдено сразу между начальным и конечным состояниями:
[$916$]U=1.5[$183$](p1V2-p1V1)=1.35 кДж
Как Вы верно определили, работа совершается только на участке 2-3
A=p2[$183$](V2-V1)=2 кДж
Q=[$916$]U+A=3.35 кДж
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