Question

The volume of a gas is 93mL when the temperature is 91 oC. If the temperature is reduced to 0 oC, what is the new volume of the gas?

a
85mL
b
70mL
c
124mL
d
93mL

Answer 1

To solve this problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature in Kelvin, provided the pressure remains constant. The formula for Charles's Law is:

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

Where:
- \( V_1 \) is the initial volume.
- \( T_1 \) is the initial temperature in Kelvin.
- \( V_2 \) is the final volume.
- \( T_2 \) is the final temperature in Kelvin.

**Step 1: Convert the temperatures from Celsius to Kelvin.**

The conversion from Celsius to Kelvin is done by adding 273.15:

- Initial temperature \( T_1 = 91 \, °C = 91 + 273.15 = 364.15 \, K \)
- Final temperature \( T_2 = 0 \, °C = 0 + 273.15 = 273.15 \, K \)

**Step 2: Use Charles's Law to solve for the new volume \( V_2 \)**

Given:
- \( V_1 = 93 \, mL \)
- \( T_1 = 364.15 \, K \)
- \( T_2 = 273.15 \, K \)

Now, we substitute these values into Charles's Law:

\[
\frac{93 \, mL}{364.15 \, K} = \frac{V_2}{273.15 \, K}
\]

Rearranging to solve for \( V_2 \):

\[
V_2 = \frac{93 \, mL \times 273.15 \, K}{364.15 \, K}
\]

**Step 3: Calculate \( V_2 \)**

\[
V_2 = \frac{93 \times 273.15}{364.15}
\]

Calculating the numerator:

\[
93 \times 273.15 = 25333.95
\]

Now, calculating \( V_2 \):

\[
V_2 = \frac{25333.95}{364.15} \approx 69.5 \, mL
\]

Thus, the new volume \( V_2 \) is approximately 70 mL when rounded to the nearest whole number.

**Final Answer:** b) 70 mL