Question

2. The temperature of the air in a room needs to be adjusted (from room temperature and pressure, 298 K and latm). The room is 10.0 m wide, 22.0 m deep, and 3.0 m high. The approximate constant pressure heat capacity of air is Cp 1.0 kJ/(K*kg)1. Assume the air in the room is an ideal gas. 2.1 What is the approximate mass of the air in the room? (Use molar volume and molar weight of air) 2.2 How much energy needs to transfer to the surrounding of the room to cool the room down by 9 degrees Fahrenheit? Assume the cooling down process is under constant pressure.

Answer 1

(2.1) The approximate mass of an air in the room which will be given as -

using an ideal gas law, we have

P V = n R T

P V = (m_air / M) R T

where, R = molar gas constant = 8.314 J/mol.K

T = temperature of an air in a room = 298 K

P = pressure of an air air in a room = 1 atm = 101325 Pa

V = voulme of a room = (10 m) x (22 m) x (3 m) = 660 m³

M = molar weight of an air = 28.97 g/mol

then, we get

[(101325 Pa) (660 m³)] = [m_air / (28.97 g/mol)] (8.314 J/mol.K) (298 K)

(6.68 x 10⁷ J) = m_air (85.5 J/g)

m_air = [(6.68 x 10⁷ J) / (85.5 J/g)]

m_air = 781286.5 g

converting g to kg :

m_air = 781.2 kg

(2.2) The energy needed for transfer to the surrounding of a room to cool which will be given by -

Q = m_air C_p$\Delta$T

Q = (781.2 kg) (1 x 10³ J/kg.K) [(298 K) - (260.3 K)]

Q = [(7.812 x 10⁵ J/K) (37.7 K)]

Q = 2.94 x 10⁷ J

Q = 29.4 MJ