Mass of air leaving a room with temp raised.

[Dantes]

A room of volume 68 m^3 contains air having an average molar mass of 36.3 g/mol. If the temperature of the room is raised from 17.2C to 25C, what mass of air will leave the room? Assume that the air pressure in the room is maintained at 43.6 kPa. Answer in units of kg.
Been messing around with the following to get it into the right form, am I on the right track?

$$P_0V = n_1RT = \frac{m_1}{M}RT_1$$

$$P_0V = n_2RT_2 = \frac{m_2}{M}RT_2$$

$$m_1-m_2 = \frac{P_0VM}{R} \times (\frac{1}{T_1}-\frac{1}{T_2})$$

 

[SpeeDFX]

Yo, that looks good to me.

 

[quicknote]

Yes, you are on the right track. To solve this problem, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. We can rearrange this equation to solve for n, the number of moles:

n = PV/RT

Since we are given the volume, temperature, and pressure of the room, we can calculate the number of moles of air in the room before and after the temperature change. We can then subtract these values to find the change in the number of moles of air. This change in moles will be equal to the moles of air that have left the room.

n1 = (43.6 kPa)(68 m^3)/(8.314 J/mol·K)(17.2+273.15 K) = 6.78 mol

n2 = (43.6 kPa)(68 m^3)/(8.314 J/mol·K)(25+273.15 K) = 7.57 mol

Change in moles = n2 - n1 = 0.79 mol

Now, we can use the average molar mass of air to convert from moles to mass:

Mass of air leaving the room = (0.79 mol)(36.3 g/mol) = 28.6 g = 0.0286 kg

Therefore, the mass of air leaving the room is 0.0286 kg.