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7. Oxygen gas is at a temperature of 40°C when it occupies a volume of 2.30 L. To what temperature should it be raised to occupy a volume of 6.50 L?

Sagot :

Sure, let's go step-by-step to solve this problem.

### Step 1: Understanding the Problem

We have:
- Initial volume, [tex]\( V_1 = 2.30 \)[/tex] liters
- Final volume, [tex]\( V_2 = 6.50 \)[/tex] liters
- Initial temperature, [tex]\( T_1 = 40^\circ \text{C} \)[/tex]

We need to find the final temperature, [tex]\( T_2 \)[/tex].

### Step 2: Convert Initial Temperature to Kelvin

First, we convert the initial temperature from Celsius to Kelvin because gas law calculations typically use Kelvin.

[tex]\[ T_1(K) = T_1(°C) + 273.15 \][/tex]
[tex]\[ T_1(K) = 40 + 273.15 \][/tex]
[tex]\[ T_1(K) = 313.15 \, \text{K} \][/tex]

### Step 3: Apply Charles's Law

Charles's Law states that for a given mass of gas at constant pressure, the volume is directly proportional to its temperature (in Kelvin).

Mathematically, it can be written as:

[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]

We need to find [tex]\( T_2 \)[/tex]. Rearranging the formula to solve for [tex]\( T_2 \)[/tex]:

[tex]\[ T_2 = \frac{V_2 \times T_1}{V_1} \][/tex]

### Step 4: Calculate the Final Temperature in Kelvin

Substitute the known values into the equation:

[tex]\[ T_2 = \frac{6.50 \, \text{L} \times 313.15 \, \text{K}}{2.30 \, \text{L}} \][/tex]
[tex]\[ T_2 = \frac{2035.475}{2.30} \][/tex]
[tex]\[ T_2 = 884.99 \, \text{K} \][/tex]

### Step 5: Convert the Final Temperature to Celsius

To convert the final temperature back to Celsius:

[tex]\[ T_2(°C) = T_2(K) - 273.15 \][/tex]
[tex]\[ T_2(°C) = 884.99 - 273.15 \][/tex]
[tex]\[ T_2(°C) = 611.84 \][/tex]

### Final Answer

The temperature should be raised to [tex]\( 611.84^\circ \text{C} \)[/tex] to occupy a volume of [tex]\( 6.50 \)[/tex] liters.
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