Question

A rubber balloon is filled with 1 L of air at 1 atm and 300 K and is then put into a cryogenic refrigerator at 100 K. The rubber remains flexible as it cools. (i) What happens to the volume of the balloon?(a) It decreases to 1/3 L.(b) It decreases to 1/√3 L.(c) It is constant.(d) It increases to √3 L.(e) It increases to 3 L.(ii) What happens to the pressure of the air in the balloon?(a) It decreases to 1/3 atm.(b) It decreases to 1/√3 atm.(c) It is constant.(d) It increases to 1/√3 atm.(e) It increases to 3 atm.

Answer 1

Answer:

The correct answers are

(a) It decreases to 1/3 L

(ii) is (c) It is constant

Explanation:

to solve this, we list out the number of knowns and unknowns so as to determine the correct equation to solve the problem

The given variables are as follows

Initial volume V1 = 1L

V2 = Unknown

Initial Temperature T1 = 300K

let us assume that the balloon is perfectly elastic

At 300K the balloon is filled and it stretches to maintain 1 atmosphere

at 100K the content of the balloon cools reducing the excitement of the gas content which also reduces the pressure, however, the balloon being perfectly elastic, contracts to maintain the 1 atmospheric pressure, hence the answer to (ii) is (c) It is constant,

For (i) since we know that the pressure of the balloon is constant

by Charles Law V1/T1 =V2/T2

or V2 = (V1/T1)×T2 =$\frac{1L}{300K}$× $100K$= $\frac{1}{3}$ × L = L/3 hence the correct answer to (i) is 1/3L