Question

A room is 16 ft x 12 ft x 12 ft. would air enter or leave the room if the temperature changed from 27°c to –3°c while the pressure remained constant? Determine the volume of the air which moved in or out of the room.

Answer 1

According to the ideal gas law, if the temperature changes while the pressure remains constant, the volume of the air in the room will also change. In this case, if the temperature changes from 27°C to -3°C, the volume of air in the room will decrease, indicating that air will leave the room.

Step-by-step explanation:

When the temperature of a gas changes, its volume also changes if the pressure remains constant. This is known as the ideal gas law. According to the ideal gas law, the volume of a gas is directly proportional to its temperature, assuming the pressure remains constant. So, when the temperature of the room changes from 27°C to -3°C, the volume of air in the room will decrease.

In order to determine the volume of air that moved in or out of the room, we need to calculate the change in volume using the ideal gas law formula. The formula is: V2 = V1 * (T2 / T1), where V1 is the initial volume, T1 is the initial temperature, V2 is the final volume, and T2 is the final temperature.

Given that the initial volume of the room is 16 ft x 12 ft x 12 ft and the initial temperature is 27°C, and the final temperature is -3°C, we can calculate the final volume of the room. Plugging in the values into the formula, we get: V2 = (16 ft x 12 ft x 12 ft) * (273 + (-3))/ (273 + 27). The final volume will be less than the initial volume, indicating that air will leave the room when the temperature decreases.

Answer 2

v= 6.52 m³

convert meter to foot we get

v=230.2 ft³

Step-by-step explanation:

Given Data

Room dimension=16ft*12ft*12ft

Temperature changed from 27°c to –3°c

To find

Volume of air

Solution

V(Volume of the room) = 16ft x 12ft x 12ft

V= 2304 ft³

convert foot to meter we get

V= 65.242 m³

R=8.205 × 10⁻⁵ m³ atm / K mol.................Gas Constant

Pressure =1 atm .......as it remained constant

Initially moles of gas = PV / RT

n= 65.242 / 8.205 × 10⁻⁵ x ( 273 + 27 )

n= 2650.50

As the temperature changes the new volume for these moles = nRT / P

m= 2650.5 x 8.205 × 10⁻⁵ x ( 273 -3) /1

m= 58.72 m³

So the difference = 65.24 m³ - 58.72 m³

v= 6.52 m³

convert meter to foot we get

v=230.2 ft³