Question

A gas has a pressure of 0.570 atm at 50.0oC. What is the pressure if the temperature drops to 37.0oC?

(1 point)
Responses

0.422 atm
0.422 atm

0.594 atm
0.594 atm

0.547 atm
0.547 atm

0.770 atm

Answer 1

To find the pressure of the gas after the temperature drops, we can use Gay-Lussac's law, which states that the pressure of a gas is directly proportional to its temperature (in Kelvin) if the volume remains constant. The formula can be expressed as:

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

Where:
- \( P_1 \) = initial pressure
- \( T_1 \) = initial temperature (in Kelvin)
- \( P_2 \) = final pressure
- \( T_2 \) = final temperature (in Kelvin)

1. First, we need to convert the temperatures from Celsius to Kelvin:
- \( T_1 = 50.0 + 273.15 = 323.15 \, K \)
- \( T_2 = 37.0 + 273.15 = 310.15 \, K \)

2. Substitute the values into the formula:
- \( P_1 = 0.570 \, \text{atm} \)
- \( T_1 = 323.15 \, K \)
- \( T_2 = 310.15 \, K \)

\[
\frac{0.570 \, \text{atm}}{323.15 \, K} = \frac{P_2}{310.15 \, K}
\]

3. Now, solve for \( P_2 \):

\[
P_2 = 0.570 \, \text{atm} \times \frac{310.15 \, K}{323.15 \, K}
\]

Calculating the right side:

\[
P_2 = 0.570 \times \frac{310.15}{323.15} \approx 0.570 \times 0.9601 \approx 0.547 \, \text{atm}
\]

Therefore, the pressure when the temperature drops to 37.0°C is approximately **0.547 atm**.

The correct response is:
**0.547 atm**.