Finding the Mass and Pressure of Air in a Room

AI Thread Summary
The discussion revolves around calculating the mass and pressure of air in a room after a temperature change. The room dimensions are 15m long, 7m wide, and 3.3m high, leading to a calculated volume of 364.5 m^3 and an initial mass of 460.5 kg of air at 11°C and 108394 Pa. The main challenge is determining the new mass and pressure when the temperature increases to 24°C. Participants clarify that in a closed system, the mass remains constant despite temperature changes. The ideal gas law (PV=nRT) is referenced, emphasizing that while the mass is unchanged, pressure may vary depending on the system's characteristics. The discussion highlights the importance of understanding the relationship between temperature, pressure, and volume in gas calculations.
adam112
Messages
5
Reaction score
0
Hi all, I don't know if this is in the right place, but anyway hopefully someone maybe able to help.

Im stuck on a couple of questions I have to answer:

A room has the dimensions 15m long, 7m wide and 3.3m high, Assuming the air in the room is 11degC, 108394Pa and has a gas constant of 0.2870 Kj/kg K, Calculate:

a) The volume of air in the room (m^3) This I have at being 364.5 m^3

b) The mass of air in the room, I have this to be 460.5kg

Now this is the part i am stuck on:

Assume now that the temperature has changed to 24degC, and the gas constant remains constant, calculate

The mass of air in the room(kg)
The pressure of air in the room (Pa)

Im not sure how to go about find these out, because to find the mass you need the pressure and vice versa( or so I understand).

If anyone can enlighten me on how to achieve the unknowns via another method it would be greatly appreciated.

Many thanks.
 
Chemistry news on Phys.org
adam112 said:
Hi all, I don't know if this is in the right place, but anyway hopefully someone maybe able to help.

Im stuck on a couple of questions I have to answer:

A room has the dimensions 15m long, 7m wide and 3.3m high, Assuming the air in the room is 11degC, 108394Pa and has a gas constant of 0.2870 Kj/kg K, Calculate:

a) The volume of air in the room (m^3) This I have at being 364.5 m^3

b) The mass of air in the room, I have this to be 460.5kg

Now this is the part i am stuck on:

Assume now that the temperature has changed to 24degC, and the gas constant remains constant, calculate

The mass of air in the room(kg)
The pressure of air in the room (Pa)

Im not sure how to go about find these out, because to find the mass you need the pressure and vice versa( or so I understand).

If anyone can enlighten me on how to achieve the unknowns via another method it would be greatly appreciated.

Many thanks.


You do?
 
Assume now that the temperature has changed to 24degC, and the gas constant remains constant, calculate

The mass of air in the room(kg)
The pressure of air in the room (Pa)

PV=nRT
if the room is a closed system, then the mass remains unchanged.
if it is not, some mass of molecules are expelled ...
 
Nice one, it makes sense now!
 
I want to test a humidity sensor with one or more saturated salt solutions. The table salt that I have on hand contains one of two anticaking agents, calcium silicate or sodium aluminosilicate. Will the presence of either of these additives (or iodine for that matter) significantly affect the equilibrium humidity? I searched and all the how-to-do-it guides did not address this question. One research paper I found reported that at 1.5% w/w calcium silicate increased the deliquescent point by...
I was introduced to the Octet Rule recently and make me wonder, why does 8 valence electrons or a full p orbital always make an element inert? What is so special with a full p orbital? Like take Calcium for an example, its outer orbital is filled but its only the s orbital thats filled so its still reactive not so much as the Alkaline metals but still pretty reactive. Can someone explain it to me? Thanks!!
I'm trying to find a cheap DIY method to etch holes of various shapes through 0.3mm Aluminium sheet using 5-10% Sodium Hydroxide. The idea is to apply a resist to the Aluminium then selectively ablate it off using a diode laser cutter and then dissolve away the Aluminium using Sodium Hydroxide. By cheap I mean resists costing say £20 in small quantities. The Internet has suggested various resists to try including... Enamel paint (only survived seconds in the NaOH!) Acrylic paint (only...
Back
Top