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 1.80 atm . 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..

Express your answer numerically in atmospheres.

Part B:

In an air-conditioned room at 19.0 ∘C, a spherical balloon had the diameter of 50.0 cm. When taken outside on a hot summer day, the balloon expanded to 51.0 cm in diameter. What was the temperature outside? Assume that the balloon is a perfect sphere and that the pressure and number of moles of air molecules remains the same.

Express your answer numerically in degrees Celsius.

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 3.70 L ? (The temperature was held constant.)

Express your answer numerically in grams.

Answer 1

Ans. #A. Boyle’s Law: For a gas at constant temperature confined in a closed vessel, the product of pressure and volume is a constant.

That is, PV = Constant                                   - Temperature kept constant’

Or, P1V1 = P2V2      - equation 1                          - T, n constant

Putting the values in equation 1-

            P1V1 (Before compression) = P2V2 (After compression)

            Or, 1.8 atm x 3.0 L = P2 x 1.0 L

            Or, P2 = (1.80 atm x 3.0 L) / 1.0 L

            Hence, P2 = 5.4 atm

Therefore, final pressure after compression = 5.4 atm

# B. Initial volume at 19.0⁰C = 4/3 (pi) (diameter / 2)³

                                                = 4/3 (pi) (50.0 cm / 2)³

                                                = 65449.85 cm³                                            ; [1L = 10³ cm³]

                                                = 65.449 L

Final volume outside room =4/3 (pi) (51.0 cm / 2)³ =69455.9 cm³ = 69.456 L

Initial temperature = 19.0⁰C = 292.15 K

# Charles’ Law: Volume (V) of a gas in a closed vessel is directly proportional to absolute temperature (T, in kelvin).

That is,            V₁ / T₁ = V₂ / T₂                   - Pressure constant            - equation 1

Note: the temperature must be in terms of “absolute temperature, K”.

Given, T1 = 19.0⁰C = 292.15 K                    ; T2 ?

Putting the values in equation 1-

            65.449 L / 292.15 K = 69.456 L / T2

            Or, T2 = 69.456 L / (65.449 L / 292.15 K)

            Hence, T2 = 310.036 K

Or, T2 = (310.036 – 273.15) K = 36.886⁰C

Therefore, outside temperature = 36.9⁰C

#C.

#C. Since the piston is moveable, the pressure in vessel would be equilibrated to the atmospheric pressure. So, pressure in vessel = 1.00 atm

            Moles of He = Mass / molar mass = 2.0 g / (4.0 g mol^-1) = 0.50 mol

            Initial volume = 2.0 L

Now, putting the values in ideal gas equation-

1.00 atm x 2.0 L = 0.50 x (0.0821 atm L mol^-1K^-1) x T

            Or, T = 2.0 atm L / (0.04105 atm L K^-1)

            Hence, T = 48.72 K

Therefore, room temperature = 48.72 K (- 224.43⁰C). Obviously -224.43⁰C is NOT a general “room temperature”. It was the source of error.

# When more He is added-

            P and T remains constant

            Final volume = 3.70 L

            Let final moles of He in vessel after addition = n

Again using ideal gas equation –

1. atm x 3.70 L = n x

Or, n = 3.70 atm L / 3.999912 atm L mol^-1

Hence, n = 0.925 mol

Total mass of He after addition = Moles x Molar mass

                        = 0.925 mol x (4.0 g/ mol) = 3.700 g

Now, mass of He added = Final Mass – Initial mass = 3.70 g – 2.00 g = 1.70 g