Question

1. the combined gas law

all pressure–volume–temperature relationships for gases can be combined into a single relationship known as the combined gas law. this expression can be used when looking at the effect of changes in two of these variables on the third as long as the amount of gas (number of moles) remains constant. to use the combined gas law properly, you must always express the temperatures in kelvins. the combined gas law can be represented as follows:
\\(\frac{p_1v_1}{t_1} = \frac{p_2v_2}{t_2}\\)

part a
a very flexible helium - filled balloon is released from the ground into the air at 20. °c. the initial volume of the balloon is 5.00 l, and the pressure is 760. mmhg. the balloon ascends to an altitude of 20 km, where the pressure is 76.0 mmhg and the temperature is - 50. °c. what is the new volume, \\(v_2\\), of the balloon in liters, assuming it doesnt break or leak?
express your answer with the appropriate units.
view available hint(s)

Explanation:

Step1: Identify known values

We know the combined gas law is \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\). From the problem:

• \(P_1 = 760\ \text{mmHg}\)
• \(V_1 = 5.00\ \text{L}\)
• \(T_1 = 20^\circ\text{C}+ 273.15=293.15\ \text{K}\) (convert Celsius to Kelvin)
• \(P_2 = 76.0\ \text{mmHg}\)
• \(T_2=- 50^\circ\text{C}+ 273.15 = 223.15\ \text{K}\)

Step2: Rearrange the formula to solve for \(V_2\)

From \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\), we can rearrange it to \(V_2=\frac{P_1V_1T_2}{P_2T_1}\)

Step3: Substitute the known values into the formula

Substitute \(P_1 = 760\ \text{mmHg}\), \(V_1 = 5.00\ \text{L}\), \(T_1 = 293.15\ \text{K}\), \(P_2 = 76.0\ \text{mmHg}\), \(T_2 = 223.15\ \text{K}\) into the formula:

$$
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Answer:

\(V_2\approx38.1\ \text{L}\)