Given 5.53 L of N2 gas at -25 °C, what volume will the nitrogen occupy at the following temperatures.?

1) 0.0 °C

2) 100. K

Answer:
Look! First you need to convert -25C to kelvin!!

C + 273.15 = K

-25C + 273.15 = 248.15K

Now convert the others for future use!

0C = 273.15K

Oh! The other one is already in K!

V1/T1 = V2/T2

Rearrange!

(V1)(T2) / (T1) = V2


1) (5.53L)(273.15K) / (248.15K) = V2

V2 = 6.087L


2) (5.53L)(100K) / 248.15K = V2

V2 = 2.228L
Do your own homework. You'll never learn anything if you just have people fill in your blanks for you!
Using PV=nRT, there are two values given and 1 constant.

So we can neglect P(pressure) and n(moles) in this. Now there are two different temperatures and volumes, we will have two equations of Vi=RTi and Vf=RTf. R is a constant, so Vi/Ti = Vf/Tf.

Now sub in the values for the formula.

Vi = Initial Volume(In liters)
Vf = Final Volume(In liters)
Ti = Initial Temperature(In Kelvin)
Tf = Final Temperature(In Kelvin)

Question 1: 5.53L / 248.15K = Vf / 273.15K
Vf = 6.09L

Question 2: 5.53L / 248.15K = Vf / 100K
Vf= 2.23L
Assuming constant pressure, 5.53 x 273/248, and 5.53 x 100/248 liters, respectively.

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