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A 4.35 L plastic bag filled with 19.8°C air is pulled down to the bottom of a 26.0 m deep lake, where the temperature is 4.00°C. What is the size of the bag there after the air in the bag cools down to the water temperature?

User Timotgl
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Answer:

The size of the bag at the bottom of the lake, after the air in the bag cools down to the water temperature, is approximately 4.113 liters.

Step-by-step explanation:

To find the size of the bag after the air inside cools down to the water temperature, we can use the ideal gas law and the principle of constant pressure. The ideal gas law is given by:

PV = nRT

Where:

P = pressure of the gas

V = volume of the gas

n = number of moles of the gas

R = ideal gas constant

T = temperature of the gas in Kelvin

Since the bag is plastic, we assume it is impermeable to water, and thus the pressure inside the bag remains constant as it descends to the bottom of the lake.

Step 1: Convert temperatures to Kelvin

T1 = 19.8°C + 273.15 = 292.95 K (initial temperature in the bag)

T2 = 4.00°C + 273.15 = 277.15 K (temperature at the bottom of the lake)

Step 2: Calculate the size of the bag at the bottom of the lake (V2).

Since the pressure (P) remains constant, we can use the following formula for constant pressure processes:

(V1/T1) = (V2/T2)

where V1 is the initial volume of the bag (4.35 L) and V2 is the volume of the bag at the bottom of the lake (unknown).

Now, rearranging the formula to solve for V2:

V2 = (V1 * T2) / T1

V2 = (4.35 * 277.15) / 292.95

V2 = 4.113 L (approximately)

So, the size of the bag at the bottom of the lake, after the air in the bag cools down to the water temperature, is approximately 4.113 liters.

User Zakiyyah
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