Textbook Question
Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is 210.0 L that contains O2 gas at a pressure of 16,500 kPa at 23 °C. (a) What mass of O2 does the tank contain?
7. Gases
The Ideal Gas Law
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A 1.20 L weather balloon on the ground has a temperature of 25.0 °C and is at atmospheric pressure (1.00 atm). When it rises to an elevation where the pressure is 0.710 atm, the new volume is 1.80 L. What is the temperature (in °C) of the air at the elevation where the balloon rises?
A
15.0 °C
B
45.0 °C
C
55.0 °C
D
35.0 °C
Verified step by step guidance1
Identify the known variables: initial volume (V1) = 1.20 L, initial temperature (T1) = 25.0 °C, initial pressure (P1) = 1.00 atm, final volume (V2) = 1.80 L, and final pressure (P2) = 0.710 atm. The final temperature (T2) is unknown.
Convert the initial temperature from Celsius to Kelvin by using the formula: T(K) = T(°C) + 273.15. This gives T1 in Kelvin.
Use the combined gas law, which is expressed as \( \frac{P1 \cdot V1}{T1} = \frac{P2 \cdot V2}{T2} \). Rearrange this equation to solve for T2: \( T2 = \frac{P2 \cdot V2 \cdot T1}{P1 \cdot V1} \).
Substitute the known values into the rearranged equation: \( T2 = \frac{0.710 \text{ atm} \cdot 1.80 \text{ L} \cdot (25.0 + 273.15) \text{ K}}{1.00 \text{ atm} \cdot 1.20 \text{ L}} \).
Calculate T2 in Kelvin and then convert it back to Celsius using the formula: T(°C) = T(K) - 273.15. This will give you the temperature of the air at the elevation where the balloon rises.
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