Question

A gas in a closed container has a pressure of 4.60 atm at a temperature of 380 K. If the pressure in the container is decreased to 3.20 atm, what is the new temperature?

A. 264 K

B. 26 K

C. 83 K

D. 546 K

Answer 1

To solve for the new temperature when the pressure of the gas changes, we can use the ideal gas law in its form that relates pressure and temperature directly, assuming the volume and the amount of gas do not change. The relationship can be given by:

\[
\frac{P_1}{T_1} = \frac{P_2}{T_2}
\]

where:
- \( P_1 = 4.60 \, \text{atm} \) (initial pressure)
- \( T_1 = 380 \, \text{K} \) (initial temperature)
- \( P_2 = 3.20 \, \text{atm} \) (final pressure)
- \( T_2 \) is the unknown final temperature

Rearranging the equation to solve for \( T_2 \):

\[
T_2 = T_1 \cdot \frac{P_2}{P_1}
\]

Now plug in the values:

\[
T_2 = 380 \, \text{K} \cdot \frac{3.20 \, \text{atm}}{4.60 \, \text{atm}}
\]

Calculating the fraction:

\[
\frac{3.20}{4.60} = 0.695652174
\]

Now, multiply:

\[
T_2 = 380 \, \text{K} \cdot 0.695652174 \approx 264 \, \text{K}
\]

Thus, the new temperature \( T_2 \) is approximately **264 K**.

The correct answer is **A. 264 K**.