7) A sample of helium has a volume of 521 dm3 at a pressure of 75 cm Hg and a temperature
of 18° C. When the temperature is increased to 23° C, what is the volume of the helium?
Answers
The volume of the helium is 530. dm³.
When you examine the question, you see that the pressure does not change. The only things that change are the volume and the temperature.
This is an example of Charles' Law.
V
1
T
1
=
V
2
T
2
V
1
= 521 dm³;
T
1
= (18 + 273.15) K = 291 K
V
2
= ?;
T
2
= (23 + 273.15) K = 296 K
V
2
=
V
1
×
T
2
T
1
= 521 dm³ ×
296
K
291
K
= 530. dm³
This makes sense. The absolute temperature increased by a factor of about 2 %, so the volume increased by about 2 % (10 dm³).
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Answer:
Assuming this sample of helium gas behaves like an "ideal gas", the equation PV=nRT describes the relationship between pressure, volume and temperature (for a given number of moles, n).
P = 75 mm Hg
V = 521 L
T = 18 C + 273 = 291 K
R = 62.4 mm Hg L/mol * K (be sure your units for this constant match the units in P, V and T)
P1 x V1 = n x R x T1 and P2 x V2 = n x R x T2
n = assume moles of gas do not change when temp increases from 18 to 23 C. Thus, n = 1 or is omitted.
P1 x V1 = R x T1 and P2 x V2 = R x T2
The problem states that volume changes with the temperature. So, we assume pressure will remain constant, i.e., Charles Law:
V1 = R x T1 and V2 = R x T2
Since R1 and R2 are the same we have a common relationship:
V1/T1 = R and V2/T2 = R
V1/T1 = V2/T2 (also known as Charles Law)
Since we are solving for V2 we isolate that term:
(V1/T1) x T2 = V2
V2 = (521 L x 296 K)/ 291 K
V2 = 530 L
Answer check: Since temperature increased 1.02 % (296K/291K), so must the volume (530 L/521 L) = 1.02 %.
EAssuming this sample of helium gas behaves like an "ideal gas", the equation PV=nRT describes the relationship between pressure, volume and temperature (for a given number of moles, n).
P = 75 mm Hg
V = 521 L
T = 18 C + 273 = 291 K
R = 62.4 mm Hg L/mol * K (be sure your units for this constant match the units in P, V and T)
P1 x V1 = n x R x T1 and P2 x V2 = n x R x T2
n = assume moles of gas do not change when temp increases from 18 to 23 C. Thus, n = 1 or is omitted.
P1 x V1 = R x T1 and P2 x V2 = R x T2
The problem states that volume changes with the temperature. So, we assume pressure will remain constant, i.e., Charles Law:
V1 = R x T1 and V2 = R x T2
Since R1 and R2 are the same we have a common relationship:
V1/T1 = R and V2/T2 = R
V1/T1 = V2/T2 (also known as Charles Law)
Since we are solving for V2 we isolate that term:
(V1/T1) x T2 = V2
V2 = (521 L x 296 K)/ 291 K
V2 = 530 L
Answer check: Since temperature increased 1.02 % (296K/291K), so must the volume (530 L/521 L) = 1.02 %.
Hope it helps you!