Question

A sample of 42 mL of carbon dioxide gas was placed in a piston in order to maintain a constant 101 kPa of pressure.

Answer 1

Answer:

The answer to your question is 33.4 ml

Explanation:

Data

volume 1 = V1 = 42 ml

temperature 1 = T1 = 20°C

temperature 2 = T2 = -60°C

Volume 2 = V2 = x

Process

1.- Convert celsius to kelvin

T1 = 20 + 273 = 293°K

T2 = -60 + 273 = 233°K

2.- Use the Charles' law to solve this problem

               $\frac{V1}{T1} = \frac{V2}{T2}$

Solve for V2

                V2 = $\frac{V1T2}{T1}$

3.- Substitution

               V2 = $\frac{(42)(233)}{293}$

4.- Simplification

                V2 = $\frac{9786}{293}$

5.- Result

                V2 = 33.4ml