Q. 8.30

Question

An air bubble has a volume of $0.500 L$ at $18 ° C$ What is the final volume, in liters, of the gas when the temperature changes to each of the following, if $P$ and $n$ do not change?
a. $0 ° C$
b. $425 K$
c. $- 12 ° C$
d. $575 K$

Step-by-Step Solution

Verified

Answer

Part(a) The final volume is $0.469 L$

Part(b) The final volume is $0.730 L$

Part(c) The final volume is $0.448 L$

Part(d) The final volume is $0.988 L$

1Part(a) Step 1 : Given information

We are given an air bubble with constant pressure and number of moles. We need to find final volume.

2Part(a) Step 2 : Simplify

As we know, $0 ° C = 273 K$

Using Charle's Law $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$

$V_{1} = 0.500 L T_{1} = 18 ° C = 18 + 273 K = 291 K T_{2} = 0 ° C = 0 + 273 K = 273 K$

Substituting the values, we get Final volume is $0.469 L$

3Part(b) Step 1 : Given information

We are given an air bubble with constant pressure and number of moles. We need to find final volume.

4Part(b) Step 2 : Simplify

As we know, $0 ° C = 273 K$

Using Charle's Law $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$

$V_{1} = 0.500 L T_{1} = 18 ° C = 18 + 273 K = 291 K T_{2} = 425 K$

Substituting the values, we get Final volume is $0.730 L$

5Part(c) Step 1 : Given information

We are given an air bubble with constant pressure and number of moles. We need to find final volume.

6Part(c) Step 2 : Simplify

As we know $0 ° C = 273 K$

Using Charle's Law $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$

$V_{1} = 0.500 L T_{1} = 18 ° C = 18 + 273 K = 291 K T_{2} = - 12 ° C = - 12 + 273 K = 261 K$

Substituting the values, we get Final volume is $0.448 L$

7Part(d) Step 1 : Given information

We are given an air bubble with constant pressure and number of moles. We need to find final volume.

8Part(d) Step 2 : Simplify

As we know, $0 ° C = 273 K$

Using Charle's Law $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$

$V_{1} = 0.500 L T_{1} = 18 ° C = 18 + 273 K = 291 K T_{2} = 575 K$

Substituting the values, we get Final volume is $0.988 L$

Q. 8.32

Q. 8.34

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Q. 8.34

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Q. 8.35

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