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?

Question

Arianna Kerr

Chemistry

7 May, 16:41

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?

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Maleah · Answer 1

0.46 L is the new volume of the balloon

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