Question

1. 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.

2. 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 ?

Answer 1

Solution:

1.Given that pressure and nmber of moles remians constant.

Volume of a cube= radius^3= (50)3=125000, (51)^3=132651, T1=190 ∘C=19+273=565

increase in volume is

V1 / T1 = V2 / T2   => T2= (V2/V1)*T1 => (132651/125000)*565=309.9K

the temperture outside the ballon =309.9K-273K= 36.9 ∘C

2.given 2 grams of helium .

n1=moles of Helium = 2/4.003 =0.5 moles

n2=moles of Helium addes => to be calculated,

V1= 2.00L

V2= 3.00L

P1V1 / n1T1 = P2V2 / n2T2       given P is constant.T is constant

V1/n1=V2/n2   > plug in the values

2/0.5 =3.70/n2

=> n2= (3.70/2*0.5) => n2= 0.925 moles

moles= mass/molar mass => mass =moles *molar mass

Mass of helium = 0.925/4.003 =0.2315 grams

2.00 g-0.2315 = 1.8 grams of Helium was added