Question

A 0.36 L balloon at 23.6 oC and 1050 mm Hg is placed in new conditions of standard temperature and pressure. What will be the new volume of the balloon?

Answer 1

0.46 L is the new volume of the balloon

Step-by-step explanation:

When the moles of a gas remain constant we have this situation for these conditions

P1 . V1 /T1 = P2 . V2/ T2

First of all we convert the mmHg to atm

1050 mmHg . 1 atm / 760 mmHg =1.38 atm

We also need absolute T°C

T° K = T°C + 273 → 23.6°C + 273 = 296.6 K

STP → 1 atm = P ; 273 K = T

Now we can replace the values:

1.38 atm . 036 L / 296.6K = 1 atm . V2 / 273K

(1.38 atm . 036 L / 296.6K) . 273 K = 1 atm . V2

( 1.38 atm . 036 L / 296.6K) . 273K) / 1 atm = V2 → 0.46 L