chem question?

Question

A sample of argon gas occupies a volume of 8.95 L at 45 oC and 0.931 atm.

If it is desired to increase the volume of the gas sample to 11.0 L, while increasing its pressure to 1.12 atm, the temperature of the gas sample at the new volume and pressure must be oC.

Answer 1

There is an easy formula you can plug all these values into:

(Pressure * Volume) /temperature = (Pressure * Volume)/temperature

or

PV/T = PV/T

the first P, V, and T are your original values (so plug in 8.95L for V, 0.931 atm for P, and 318.15 K for T *you must convert your 45 oC to K; to do this you must add 273.15 to your temp, here it's 273.15 + 45*) ---- if you do his you'll get

(0.931 atm)(8.95 L)/(318.15K) = PV/T

for the second P, V, and T, you plug in the values that you already know of what you want to get (so plug in 1.12 atm for P, and plug in 11.0L for V) now you're left with

(0.931 atm)(8.95 L)/(318.15K) = (1.12 atm)(11.0L)/T

solve for T

T (0.931 atm)(8.95 L)/(318.15K) = (1.12 atm)(11.0L)

T = [(1.12 atm)(11.0L)(318.15 K)] / [(0.931 atm)(8.95 L)]

T = 470 K

now convert to oC by subtracting 273.15

T = 470 K = 197 oC

and there you go. same steps for all these types of problems, just maybe solve for different variable depending on what kind of data they give you

Answer 2

what's the rule here, Boyle's Law? PV = nrT? let's simplify to PV = kT.

(0.931 atm)(8.95 L) = k(45° C)
------------------------- ... ------------
(1.12 atm)(11.0 L) ..=..... kT

k's cancel, leaving a proportion which solves as
T = (45)(1.12)(11.0)/[(0.931)(8.95)]
T = 66.535° C