Part A A certain amount of chlorine gas was placed inside a cylinder with a mova

Question

Part A

A certain amount of chlorine gas was placed inside a cylinder with a movable piston at one end. The initial volume was 3.00 L and the initial pressure of chlorine was 1.95atm. The piston was pushed down to change the volume to 1.00 L. Calculate the final pressure of the gas if the temperature and number of moles of chlorine remain constant..

Part B

Part C

A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 4.50L ? (The temperature was held constant.)

Part D

If a certain gas occupies a volume of 15L when the applied pressure is 7.5 atm , find the pressure when the gas occupies a volume of 3.8L .

Explanation / Answer

for question A & B)

Boyle's law states that (P1)(V1)=(P2)(V2), if temperature and moles are held constant. Ergo, (1.25 atm)(3.00 L)=(P2)(1.00 L). Consequently, P2=3.75 atm.

Re: spherical balloon.
Charles' law states that (V1)/(T1)=(V2)/(T2), if pressure and moles are held constant, and with temperature in an absolute scale like Kelvin.

V1=(4/3)(50.0/2)^3*pi=20833.3 cm^3
V2=(4/3)(51.0/2)^3*pi=22108.5 cm^3
T1=19.0+273.15=292.15 K

(20833.3)/(292.15)=(22108.5)/(T2)
T2=310.0 K=36.88 degrees C.

C &D)

P x V = n x R x T

n = m/MW

==> P x V = (m/MW) x R x T

Solving for (R / MW)

==> R / MW = (P x V) / (m x T)
R / MW = constant for all states

So, at states 1 and 2,
(P1 x V1) / (m1 x T1) = (P2 x V2) / (m2 x T2)

Given: P1 = P2 and T1 = T2 (both constant)

==> V1 / m1 = V2 / m2

Solving for m2,
m2 = (V2 x m1) / V1

Given:
V1 = 2.00L
m1 = 2.00 g
V2 = 4.30L

Substituting values,

m2 = (4.30L x 2.00 g) / 2.00L

= 4.30 g

The mass added = m2 - m1

= 4.30 g - 2.00 g

= 2.30 g of He

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