Multiple Choice
What is the volume of 1.00 mole of an ideal gas at 1.00 atm pressure and 0°C?
7. Gases
The Ideal Gas Law
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What temperature (in °C) did an ideal gas shift to if it was initially at -18.0 °C at 4.62 atm and 35.0 L, and the pressure was changed to 8.31 atm and the volume changed to 25.0 L?
A
27.0 °C
B
0.0 °C
C
45.0 °C
D
15.0 °C
Verified step by step guidance1
Identify the initial and final conditions of the gas: initial temperature (T1) = -18.0 °C, initial pressure (P1) = 4.62 atm, initial volume (V1) = 35.0 L, final pressure (P2) = 8.31 atm, and final volume (V2) = 25.0 L.
Convert the initial temperature from Celsius to Kelvin, since gas law calculations require temperatures in Kelvin. Use the formula: T(K) = T(°C) + 273.15. So, T1(K) = -18.0 + 273.15.
Apply the combined gas law, which relates the pressure, volume, and temperature of a gas: \( \frac{P1 \cdot V1}{T1} = \frac{P2 \cdot V2}{T2} \). Rearrange this equation to solve for the final temperature (T2): \( T2 = \frac{P2 \cdot V2 \cdot T1}{P1 \cdot V1} \).
Substitute the known values into the rearranged equation: \( T2 = \frac{8.31 \text{ atm} \cdot 25.0 \text{ L} \cdot T1(K)}{4.62 \text{ atm} \cdot 35.0 \text{ L}} \).
Calculate T2 in Kelvin using the substituted values, then convert T2 back to Celsius using the formula: T(°C) = T(K) - 273.15.
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