Mass of air leaving a room with temp raised.

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The discussion focuses on calculating the mass of air leaving a room when the temperature is raised from 17.2°C to 25°C, with a room volume of 68 m³ and air pressure maintained at 43.6 kPa. The molar mass of the air is given as 36.3 g/mol. The formula used for the calculation is derived from the ideal gas law, specifically P_0V = nRT, leading to the conclusion that the mass difference can be expressed as m_1 - m_2 = (P_0VM/R) × (1/T_1 - 1/T_2).

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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})
 
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