Q. 8.29

Question

A balloon contains $2500 mL$ of helium gas at $75 ° C$ . What is the final volume, in milliliters, of the gas when the temperature changes to each of the following, if $P$ and n do not change?

a. $55 ° C$

b. $680 . K$

c. $- 25 ° C$

d. $240 . K$

Step-by-Step Solution

Verified

Answer

The final volume, in milliliters, of the gas when the temperature changes to each of the following, if P and n do not change :

a. $1833.3 mL$

b. $13566.6 mL$

c. $- 832.5 mL$

d. $- 1100 mL$

1Part(a) Step 1: Given information

We need to find that final volume, in milliliters, of the gas when the temperature changes to each of the following, if Pressure and amount do not change.

2Part(a) Step 2: Explanation

By using charles law , we get ;

$V_{2} = \frac{V_{1} \times T_{2}}{T_{1}} = \frac{2500 \times 55}{75} = 1833.3 mL$

3Part(b) Step 1: Given information

We need to find that final volume, in milliliters, of the gas when the temperature changes to each of the following, if Pressure and amount do not change.

4Part(b) Step 2: Explanation

By using charles law , we get ;

We will convert the given temperature in celcius ,

$T_{2} = 680 - 273 = 407 ° C$

$V_{2} = \frac{V_{1} \times T_{2}}{T_{1}} = \frac{2500 \times 407}{75} = 13566.6 mL$

5Part(c) Step 1: Given information

We need to find that final volume, in milliliters, of the gas when the temperature changes to each of the following, if Pressure and amount do not change.

6Part(c) Step 2: Explanation

By using charles law , we get ;

$V_{2} = \frac{V_{1} \times T_{2}}{T_{1}} = \frac{2500 \times (- 25)}{75} = - 832.5 mL$

7Part(d) Step 1: Given information

We need to find that final volume, in milliliters, of the gas when the temperature changes to each of the following, if Pressure and amount do not change.

8Part(d) Step 2:Explanation

By using Charles law, we get ;

We will give temperature from Kelvin to Celcius

$T_{2} = 240 - 273 = - 33 ° C$

$V_{2} = \frac{V_{1} \times T_{2}}{T_{1}} = \frac{2500 \times (- 33)}{75} = - 1100 mL$