The Ideal gas law?

Question

Okay so I have tried to answer this problem, but I am obviouslly not getting the correct answer.

Suppose a 28.1-mL sample of helium gas at 25 °C and 1.29 atm is heated to 90. °C and compressed to a volume of 15.2 mL. What will be the pressure of the sample?

V1=28.1 ml or .0281L
P1= 1.68 atm
T1= 298K
n1=1.930 mol

V2= 15.2 mL or .0152L
P2= ?
T2= 298K
N2= 1.930 mol

P2=P1XV1/V2

which is 1.29X.0281L/.0152L
= 2.38480 atm

Can someone let me know if I am doing something wrong. I would greatly appreciate it.

Can one of you let me know if I used the right formula and why I had to incorporate the temp difference. I just want to know how you got the answer. Thank you.

Thank you very much. I now understand why the temp has to be taken into consideration.

Answer 1

Since the same sample of the gas heated and compressed, the number of moles of the gas remains constans. Since the pressure of a gas is inversely proportional to its volume and directly proportional to its temperature;

P1V1/T1 = P2V2/T2

P2 = P1 x V1 / V2 x T2 / T1

P1 = 1.29 atm
V1 = 28.1 mL
V2 = 15.2 mL
T1 = 25 + 273 = 298 K
T2 = 90 + 273 = 363 K

P2 = 1.29 atm x 28.1 mL / 15.2 mL x 363 K / 298 K
P2 = 2.90 atm

Note that; in this type of comparison problems you can use mL instead of L, because the units are cancelled. But you can not do the same thing for temperature. Because, in mL to L conversion, Both V1 and V2 are divided by the same factor, their effects are cancelled. But in temperature conversion from C to K , you add (not multiply) 273. Their effects are not cancelled.

28.1 mL / 15.2 mL = 1.85
0.0281 L / 0.0152 L = 1.85

363 K / 298 K = 1.22
90 C / 25 C = 3.6

Did you notice the difference?

Answer 2

oh my dear the T2 value is 273+90=363 kelvin but not 298 as u hv written..... and the answer for that is 2.904

Answer 3

Look at T2 again.

P1V1/T1 = P2V2/T2