Какая масса m воздуха выйдет из комнаты объемом 60 м3 при повышении температуры от T1=280 до Т2=300К при нормальном давлении. Решие пожалуйста,буду очень благодарна!!!!

Какая масса m воздуха выйдет из комнаты объемом 60 м3 при повышении температуры от T1=280 до Т2=300К при нормальном давлении. Решие пожалуйста,буду очень благодарна!!!!
- Предмет:
  Физика
- Автор:
  benjaminclark
- 5 лет назад

Ответы 1

даноV = 60м3T1=280КТ2=300КR =8.31 Дж/мольКpo = 10^5 Па при нормальном давлении.M = 0.029 кг/мольнайти∆mрешениесостояние 1poV = m1/M RT1 ; m1 = MpoV/RT1состояние 1poV = m2/M RT2 ; m2 = MpoV/RT2∆m = m1-m2 = MpoV/RT1 - MpoV/RT2 = MpoV/R *(1/T1 -1/T2)∆m = 0.029*10^5*60/8.31 *(1/280 -1/300)= 4.985 кгответ4.985 кг
- Автор:
  daugherty
- 5 лет назад
- 0
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Какая масса m воздуха выйдет из комнаты объемом 60 м3 при повышении температуры от T1=280 до Т2=300К при нормальном давлении. Решие пожалуйста,буду очень благодарна!!!!

Ответы 1

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