To find the new pressure of the gas when the temperature drops, we can use the ideal gas law (assuming the volume and the amount of gas remain constant) and the relationship given by the ideal gas law derived equation:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Where:
- \( P_1 = 0.570 , \text{atm} \) (initial pressure)
- \( T_1 \) = initial temperature in Kelvins
- \( P_2 \) = final pressure (unknown)
- \( T_2 \) = final temperature in Kelvins
First, we need to convert the temperatures from Celsius to Kelvin:
-
For \( T_1 \): \[ T_1 = 50.0 , \text{oC} + 273 = 323.0 , K \]
-
For \( T_2 \): \[ T_2 = 37.0 , \text{oC} + 273 = 310.0 , K \]
Now, we can rearrange the ideal gas law equation to solve for \( P_2 \): \[ P_2 = P_1 \cdot \frac{T_2}{T_1} \]
Substituting the values: \[ P_2 = 0.570 , \text{atm} \cdot \frac{310.0 , K}{323.0 , K} \]
Calculating \( P_2 \): \[ P_2 = 0.570 \cdot \frac{310.0}{323.0} \approx 0.570 \cdot 0.9607 \approx 0.547 , \text{atm} \]
Therefore, the pressure when the temperature drops to 37.0°C is approximately 0.547 atm.
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