Question

Can someone please help me with this? I need equation help.

Answer 1

If a 10. m^3 volume of air (acting as an ideal gas) is at a pressure of 760 mm and a temperature of 27 degrees Celsius is taken to a high altitude where the pressure is 400. mm Hg and a temperature of -23 degrees Celsius, what volume will it occupy? (Hint: remember the temperature must be in Kelvin).

Answer 2

This is the question. The formulas I was given are:

Answer 3

\[P_{1}V _{1} = P_{2} V_{2}\]

Answer 4

\[V_{1}/T_{1} = V_{2}/T_{2}\]

Answer 5

\[P_{1} V_{1}/T_{1} = P_{2} V_{2}/ T_{2}\]

Answer 6

I have no idea which formula to use, or if I need to make conversions from the 400 mm Hg

Answer 7

@Robert136 could you maybe look? I would appreciate it so much

Answer 8

@Robert136 could you maybe look? I would appreciate it so much

Answer 9

@Elsa213

Answer 10

The question is poorly worded. Try again.

Answer 11

I know.. I'll give you a medal anyway since you were kind enough to reply

Answer 12

I get it now.

Answer 13

1.Assume that there is 10m^3 volume of air
2.The air above has a pressure of 760mm & temperature of 300 Kelvin degrees
3.The air mentioned above is taken to where pressure is 400mm and 250 Kelvin
4. Calculate the volume that results from change in temperature and pressure.

Answer 14

Ah I see now. So I need to plug those values into the last equation I posted and solve?

Answer 15

Definitely!

Answer 16

Thank you so so much! That cleared up a ton of confusion! I appreciate it! :D

Answer 17

No problem. I am assuming this is a last year high school chemistry or first year university?

Answer 18

I had to do it a while ago

Answer 19

It's actually for Physics. They taught me how to use the formulas but threw this curveball at me, and I had no idea what formula would work or if I needed to do conversions for it beforehand.

Answer 20

Oh I see your problem now. Make sure to use the Kelvin. Usually units of the pressure doesn't matter because they are all proportional .

Answer 21

for an ideal gas, you will have been given something like

\(\huge pV= nRT\)

as R = const, we can say:

\(\huge \frac{pV}{nT} = const\)

as it is the same amount of gas, we can say


\(\huge \frac{pV}{T} = const\)

but as @Robert136 so expertly points out, you need to be sure that you use Kelvin

and you are solving \(\huge \frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}\)

thus

\(\huge V_2 = \frac{T_2}{T_1} . \frac{ p_1 }{ p_2}\ . V_1\)

again, using Kelvin

https://gyazo.com/dad5d1f102f61add6b05a906ef9a617f

Answer 22

@IrishBoy123 thank you for your response! :D