A gas balloon has a volume of 106.0 liters when the temperature is 45.0 °C and the pressure is 740.0 mm of mercury. What will its volume be at 20.0 °C and 780 .0 mm of mercury pressure
Use (P1V1/T1) = (P2V2/T2)
Remember T must be in kelvin. K = 273 + C = ?
Post your work if you get stuck.
To solve this problem, we'll need to use the combined gas law, which states that the ratio of the initial pressure, volume, and temperature of a gas to its final pressure, volume, and temperature must be equal.
The combined gas law formula is:
(P1 * V1) / T1 = (P2 * V2) / T2
In this case, we want to find the volume at 20.0 °C and 780.0 mm of mercury pressure (V2), and we know the initial volume is 106.0 liters, initial temperature is 45.0 °C, initial pressure is 740.0 mm of mercury, and final temperature is 20.0 °C (T1, P1, and T2 are given).
First, we need to convert the pressure from mm of mercury to atm:
1 atm = 760 mmHg
So, the initial pressure in atm is 740.0 mmHg /760 mmHg = 0.974 atm.
The final pressure in atm is 780.0 mmHg/760 mmHg = 1.026 atm.
Now we can plug in the values into the combined gas law equation:
(0.974 atm * 106.0 L) / 318.0 K = (1.026 atm * V2) / 293.0 K
Simplifying the equation:
101.2 atm * L / K = 1.026 atm * V2 / K
Now solve for V2, which is the volume at 20.0 °C and 780.0 mm of mercury pressure:
V2 = (101.2 atm * 106.0 L * 293.0 K) / (1.026 atm * 318.0 K)
V2 = (3216410.4 atm * L * K) / (329066.8 atm * K)
V2 = 9.78 L
Therefore, the volume of the gas balloon at 20.0 °C and 780.0 mm of mercury pressure will be 9.78 liters.