Question

At 90.0C, XeFe4 is a gas. When 0.394g of this noble-gas compound is introduced into a 0.91 L gas bulb at 90.0C, the pressure is 47.3 mm Hg. What pressure would this same amount of gas exhibit when it is transferred to a 2.13 L bulb at 120.0C?

Answer 1

we know that

for gases

PV = nRT

now

given that

the amount of gas remains constant

so

n = constant

also

R = universal gas constant

so

PV/ T = nR = constant

so

PV/T = constant

now

for two different conditions

P1V1/T1 = P2V2 / T2

so

P2 = ( P1V1) ( T2) / ( T1V2)

given

P1 = 47.3 mm Hg

V1 = 0.91 L

T1 = 90 C = 90 + 273 K = 363 K

P2 = ??

V2 = 2.13 K

T2 = 120 C = 120 + 273 = 393 K

now

using those values , we get

P2 = ( 47.3 x 0.91) ( 393) / ( 363 x 2.13)

P2 = 21.88 mm Hg

so

the new pressure the gas exhibits is 21.88 mm Hg