Question

Consider 4.60 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.60 L and the temperature is increased to 36 ∘C , what is the new pressure, P2, inside the container? Assume no change in the amount of gas inside the cylinder

Answer 1

Answer:

P2 = 681.03 mmHg

Explanation:

Before moving any further, let's bring out all the parameters stated in the question.

Initial Volume (V1) = 4.60L

Initial Pressure (P1) = 365 mmHg

Initial Temperature (T1) = 20 C + 273 = 293K (Converting to Kelvin Temperature)

Final Temperature (T2) = 36 C + 273 = 309K (Converting to Kelvin Temperature)

Final Volume (V2)  = 2.60 L

Final Pressure (P2) = ?

The problem stated three gas parameters (volume, pressure and temperature), hence we are going to use the general gas equation because it is the formular that defines the relationship between the three gas parameters.

It is given as;

$\frac{P1V1}{T1} = \frac{P2V2}{T2}$

Making P2 subject of interest, we are left with;

$P2 = \frac{P1V1T2}{V2T1}$

P2 = (365 * 4.6 * 309) / (2.6 * 293)

P2 = 681.03 mmHg

Answer 2

Answer:

P=681.07mmHg

Explanation:

gas law:

(V1 * P1 ) / T1 = (V2 * P2 ) / T2

⇒ P2 = ( V1 * P1 * T2) / ( T1 * V2)

∴ V1 = 4.6L;

  P1 = 365mmHg * 0.001315 atm / 1 mmHg = 0.48 atm

  T2 = 36°C + 273 = 309 K

  T1 = 20 °C + 273 = 293 K

  V2 = 2.6 L

⇒ P2 = ((4.6 L) * (0.48 atm) * (309 K)) / ((293 K) * (2.6 L))

⇒P2 = 0.895 atm ( 681.07 mmHg)

when compressing the gas inside the cylinder, the internal pressure increases