Определите работу и мощность тока в электрической лампе, если при напряжении 3 вольта сила тока в ней была равна 100 мампер, лампа работала 5 минут.

Мощность P=U*I; P=3*0,1=0,3Вт
Работа A=P*t; A=0,3*(5*60)=90Дж.

T=5m = 5*60=300с
A=IUt = 3*100*10^-3 * 300=90 Дж 
N=P=A/t  = 90 / 300 = 0.3 Вт

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на протяжении всей операции не меняет свой объем.

Выразим V из закона Менделеева-Клапейрона: 

P V = m R T / M => V = m R T / P M.

А теперь приравняем V1 к V2. И дабы не писать лишнего, сразу посмотрим, что у нас сократится: M, R, m (но сначала я напишу с m для ясности). Получаем:

m T1 / P1 = 0,4 m T2 / P2.

У тебя сейчас, наверное, возник вопрос: почему во второй части уравнения перед m стоит 0,4?

- Потому что исходя из условия задачи мы можем сделать вывод, что m2 = 0,4 m1 (в уравнении m1 заменена на просто m для краткости). 

Теперь сокращаем массы, выводим P2:

P2 = 0,4 T2 P1 / T1 = 4*10^-1 * 273 * 2*10^5 / 3*10^2 = 72,8*10^3 Па