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 doesn't break or leak?

To find the new volume of the balloon, we can use the combined gas law equation:

(P1 * V1) / T1 = (P2 * V2) / T2

where:
P1 = initial pressure = 760 mmHg
V1 = initial volume = 5.00 L
T1 = initial temperature = 20 °C + 273.15 = 293.15 K
P2 = final pressure = 76.0 mmHg
T2 = final temperature = -50 °C + 273.15 = 223.15 K
V2 = unknown

Let's substitute the given values into the equation and solve for V2:

(760 mmHg * 5.00 L) / 293.15 K = (76.0 mmHg * V2) / 223.15 K

3800 mmHg L / 293.15 K = 76.0 mmHg * V2 / 223.15 K

Cross-multiplying and simplifying:

76.0 mmHg * V2 = (3800 mmHg L * 223.15 K) / 293.15 K

76.0 mmHg * V2 = 794310 mmHg L / K

Dividing both sides by 76.0 mmHg:

V2 = (794310 mmHg L / K) / 76.0 mmHg

V2 ≈ 10450.9 L

Therefore, the new volume of the balloon at an altitude of 20 km is approximately 10450.9 liters.

To solve this problem, we can use the combined gas law equation, which combines Boyle's Law, Charles's Law, and Gay-Lussac's Law.

The combined gas law equation is given as:
(P1 * V1) / (T1) = (P2 * V2) / (T2)

Where:
P1 and P2 are the initial and final pressure, respectively,
V1 and V2 are the initial and final volume, respectively,
and T1 and T2 are the initial and final temperature, respectively.

In this problem, we are given:
P1 = 760 mmHg (initial pressure),
V1 = 5.00 L (initial volume),
T1 = 20 °C (initial temperature),
P2 = 76.0 mmHg (final pressure),
T2 = -50 °C (final temperature).

Before using the equation, we need to convert the temperature from Celsius to Kelvin because the equation requires temperatures in Kelvin scale. To convert Celsius to Kelvin, we use the following formula:
T(K) = T(C) + 273.15

Converting the temperatures:
T1(K) = 20 + 273.15 = 293.15 K
T2(K) = -50 + 273.15 = 223.15 K

Now, substituting the given values into the combined gas law equation:
(760 mmHg * 5.00 L) / (293.15 K) = (76.0 mmHg * V2) / (223.15 K)

To solve for V2, we rearrange the equation:
V2 = (760 mmHg * 5.00 L * 223.15 K) / (76.0 mmHg * 293.15 K)

Simplifying the equation:
V2 = (3809738 mmHg * L * K) / 221792.2 mmHg K

Finally, converting mmHg to atm:
1 atm = 760 mmHg
V2 = (3809738 mmHg * L * K) / (221792.2 mmHg K) * (1 atm / 760 mmHg)

Simplifying the final equation gives us the value of V2 in liters.

If you can give us some hints as to what you don't understand about all of these questions someone here may be able to help. You can google most of the answers.

Wow! Nine posts in four minutes without a single thought of your own! You must be taking a test!!