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?

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 absolute temperature (in Kelvin) when the pressure is constant. Charles's Law is given by the formula:

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

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.

1. **Convert temperatures from Celsius to Kelvin**:
- \( T_1 = 91 \, ^\circ C + 273.15 = 364.15 \, K \)
- \( T_2 = 0 \, ^\circ C + 273.15 = 273.15 \, K \)

2. **Apply Charles's Law**:
- \( V_1 = 93 \, mL \)
- Using the values in the formula:

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

3. **Cross-multiply to solve for \( V_2 \)**:

\[
93 \, mL \cdot 273.15 \, K = V_2 \cdot 364.15 \, K
\]

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

4. **Calculate \( V_2 \)**:

\[
V_2 = \frac{25410.95 \, mL \cdot K}{364.15 \, K} \approx 69.8 \, mL
\]

Thus, the new volume of the gas when the temperature is reduced to \( 0 \, ^\circ C \) is approximately **69.8 mL**.