дано
v=1 моль
T1=600K
P1=4*10^5 Па
T1/T2= V2/V1
P2=10^5 Па
Найти ∆U
Решение
P1V1 = vRT1 ; V1 = vRT1/P1 (1)
P2V2 = vRT2; V2 = vRT1/P2 (2)
Делим (2) на (1)
V2/V1 = vRT2/P2 / vRT1/P1 = T2/T1 * P1/P2 (3)
По условию T1/T2= V2/V1 тогда (3) запишем в виде
T1/T2 = T2/T1 * P1/P2
(T1/T2)^2 = P1/P2
Подставим значения
(600/T2)^2 = 4*10^5 / 10^5 = 4 = 2^2
600/T2 = 2
T2 =300 K
Аргон - одноатомный газ
Изменение внутренней энергии одноатомного газа
∆U = 3/2 *vR∆T = 3/2 *1*8.31*(300-600) = - 3739,5 Дж
охлаждается T2 < T1 - ответ с МИНУСОМ
ответ 3739,5 Дж = 3,7 кДж
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Объяснение:
Эта задача обратна той, которая находится во вложении!