Question

I have 140 L of a gas in a piston at a temperature of 523 K. If I cool the gas until the volume decreases to 65 L, what will the temperature of the gas now be?(1 point)
Responses

-30.2 K
-30.2 K

861 K
861 K

343 K
343 K

243 K

Answer 1

To determine the new temperature of the gas when its volume changes, we can use the Charles's Law, which relates the volume and temperature of a gas at constant pressure. Charles's Law states that:

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

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).

We are given:
- \( V_1 = 140 \, \text{L} \)
- \( T_1 = 523 \, \text{K} \)
- \( V_2 = 65 \, \text{L} \)

We are to find \( T_2 \). Rearranging the equation gives:

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

Substituting in the known values:

\[
T_2 = 523 \, \text{K} \times \frac{65 \, \text{L}}{140 \, \text{L}}
\]

Calculating \( \frac{65}{140} \):

\[
\frac{65}{140} = 0.4643
\]

Now, substitute back into the equation for \( T_2 \):

\[
T_2 = 523 \, \text{K} \times 0.4643 \approx 242.57 \, \text{K}
\]

Rounding this, we can conclude:

\[
T_2 \approx 243 \, \text{K}
\]

Therefore, the answer is **243 K**.