Solution to y11 q2 gases (a) At 20oC V1 =6000cm3 P constant

At 70oC V2 =? Use the combined gas equation with P canceling

out

2

22

1

11

T

VP

T

VP

2

2

1

1

T

V

T

V

Remember to convert T Temperature is always in Kelvin in the gas law

equations. Kelvin = oC+273 200C = 293K 70oC = 343K

343

volumenew

293

6000

32 7024

293

3436000cmV

(b) The expansion is the same, so the volume of gas

that escapes when heated is 7024-6000 = 1024cm3

as a fraction, this is the same as the fraction of mass that has escaped, and is:

146.07024

1024

(c)

(i) mass of cold air = volume x density Mass = 250 x 1.3 = 325 kg

(ii) mass of hot air = 250 x 0.975 = 244 kg (the rest has escaped)

(iii) upward force = Wcold – Whot Weight = mass in kg x 10 So upward force = 3250 – 2440 = 810N

(d) (i) the less dense air inside the balloon has

fewer molecules in each cm3 but it is able to exert a pressure equal to the denser cold air outside because the molecules are moving faster.

Each collision with the inside of the balloon gives more force.

(ii) the air inside will cool down because it will pass its thermal (motion) energy to the cooler air outside, through the balloon.