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 1.40 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.

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 in degrees Celsius? Assume that the balloon is a perfect sphere and that the pressure and number of moles of air molecules remains the same.

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.)

Answer 1

a)Since temperature and moles of chlorine remain constant gas law equation PV=nRT

gas law equation becomes P1V1= P2V2( since RHS, nRT remains constant)

P1= 1.4 atm , V1= 3L , V2=1L, P2=?

P2= P1V1/V2= 1.4*3/1 =4.2 atm

b) when pressure and number of moles remain constant, PV= nRT becones P is proportional to T

V1/T1= V2/T2

T1= 19 deg.c= 19+273.15= 292.15K

V1^{= (}4/3)*3.412*(50/2)³ cm3    V2= (4/3)*3.412*(51/2)³

T2=V2*T1/V1= (51/50)³ *292.15=310.13K

c) from PV=nRT

moles= mass/Molecular weight =2/4 =0.5

V₁/n₁= V₂/n₂ ( since P and T remained constant)

n2= V2*n1/V1= (3.7/2)*0.5 =0.925 moles

Mass of helium =0.925*4= 3.7 gms