To solve this problem, we need to use the ideal gas law, which states that the pressure of a gas times its volume is equal to the number of moles of the gas times the gas constant times the temperature (PV=nRT). However, in this case, we are not given the temperature or the number of moles of nitrogen gas, so we need to make some assumptions.

First, let's assume that the temperature is at standard room temperature, which is approximately 298 Kelvin.

Second, we can calculate the volume of the gas cylinder using the formula for the volume of a cylinder, which is πr²h. The radius of the cylinder is half of the diameter, so r = 6.9cm / 2 = 3.45cm. The height of the cylinder is 28cm. Therefore, the volume of the cylinder is π(3.45cm)²(28cm) = 1033.6 cm³.

We need to convert this volume to liters, since the volume of a gas is usually expressed in liters. There are 1000 cm³ in a liter, so the volume of the cylinder is 1033.6 cm³ / 1000 = 1.0336 liters.

Next, we can use the ideal gas law to calculate the number of moles of nitrogen gas in the cylinder. Rearranging the formula to solve for n, we get n = PV/RT. The pressure of the gas is 24.3 atm, the volume is 1.0336 liters, the gas constant R is 0.0821 L·atm/K·mol, and the temperature is 298 K. Plugging these values into the formula, we get n = (24.3 atm)(1.0336 L) / (0.0821 L·atm/K·mol)(298 K) = 1.02 moles.

Finally, we can use the ideal gas law again to calculate the final volume of the nitrogen gas when it is allowed to escape into the plastic bag. This time, we are solving for V, so the formula is V = nRT/P. The number of moles is 1.02, the gas constant is 0.0821 L·atm/K·mol, the temperature is 298 K, and the pressure is 1 atm (since the gas is now at atmospheric pressure). Plugging these values into the formula, we get V = (1.02 mol)(0.0821 L·atm/K·mol)(298 K) / 1 atm = 24.9 liters.

Therefore, the final volume of the nitrogen gas, including what's collected in the plastic bag and what's left over in the cylinder, is 24.9 liters. This answer has three significant digits, which is appropriate given the number of significant digits in the original problem.

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To solve this problem, we need to use the ideal gas law, which states that the pressure of a gas times its volume is equal to the number of moles of the gas times the gas constant times the temperature (PV=nRT). However, in this case, we are not given the temperature or the number of moles of nitrogen gas, so we need to make some assumptions.

First, let's assume that the temperature is at standard room temperature, which is approximately 298 Kelvin.

Second, we can calculate the volume of the gas cylinder using the formula for the volume of a cylinder, which is πr²h. The radius of the cylinder is half of the diameter, so r = 6.9cm / 2 = 3.45cm. The height of the cylinder is 28cm. Therefore, the volume of the cylinder is π(3.45cm)²(28cm) = 1033.6 cm³.

We need to convert this volume to liters, since the volume of a gas is usually expressed in liters. There are 1000 cm³ in a liter, so the volume of the cylinder is 1033.6 cm³ / 1000 = 1.0336 liters.

Next, we can use the ideal gas law to calculate the number of moles of nitrogen gas in the cylinder. Rearranging the formula to solve for n, we get n = PV/RT. The pressure of the gas is 24.3 atm, the volume is 1.0336 liters, the gas constant R is 0.0821 L·atm/K·mol, and the temperature is 298 K. Plugging these values into the formula, we get n = (24.3 atm)(1.0336 L) / (0.0821 L·atm/K·mol)(298 K) = 1.02 moles.

Finally, we can use the ideal gas law again to calculate the final volume of the nitrogen gas when it is allowed to escape into the plastic bag. This time, we are solving for V, so the formula is V = nRT/P. The number of moles is 1.02, the gas constant is 0.0821 L·atm/K·mol, the temperature is 298 K, and the pressure is 1 atm (since the gas is now at atmospheric pressure). Plugging these values into the formula, we get V = (1.02 mol)(0.0821 L·atm/K·mol)(298 K) / 1 atm = 24.9 liters.

Therefore, the final volume of the nitrogen gas, including what's collected in the plastic bag and what's left over in the cylinder, is 24.9 liters. This answer has three significant digits, which is appropriate given the number of significant digits in the original problem.

Knowee AI · Accepted Answer

To solve this problem, we need to use the ideal gas law, which states that the pressure of a gas times its volume is equal to the number of moles of the gas times the gas constant times the temperature (PV=nRT). However, in this case, we are not given the temperature or the number of moles of nitrogen gas, so we need to make some assumptions.

First, let's assume that the temperature is at standard room temperature, which is approximately 298 Kelvin.

Second, we can calculate the volume of the gas cylinder using the formula for the volume of a cylinder, which is πr²h. The radius of the cylinder is half of the diameter, so r = 6.9cm / 2 = 3.45cm. The height of the cylinder is 28cm. Therefore, the volume of the cylinder is π(3.45cm)²(28cm) = 1033.6 cm³.

We need to convert this volume to liters, since the volume of a gas is usually expressed in liters. There are 1000 cm³ in a liter, so the volume of the cylinder is 1033.6 cm³ / 1000 = 1.0336 liters.

Next, we can use the ideal gas law to calculate the number of moles of nitrogen gas in the cylinder. Rearranging the formula to solve for n, we get n = PV/RT. The pressure of the gas is 24.3 atm, the volume is 1.0336 liters, the gas constant R is 0.0821 L·atm/K·mol, and the temperature is 298 K. Plugging these values into the formula, we get n = (24.3 atm)(1.0336 L) / (0.0821 L·atm/K·mol)(298 K) = 1.02 moles.

Finally, we can use the ideal gas law again to calculate the final volume of the nitrogen gas when it is allowed to escape into the plastic bag. This time, we are solving for V, so the formula is V = nRT/P. The number of moles is 1.02, the gas constant is 0.0821 L·atm/K·mol, the temperature is 298 K, and the pressure is 1 atm (since the gas is now at atmospheric pressure). Plugging these values into the formula, we get V = (1.02 mol)(0.0821 L·atm/K·mol)(298 K) / 1 atm = 24.9 liters.

Therefore, the final volume of the nitrogen gas, including what's collected in the plastic bag and what's left over in the cylinder, is 24.9 liters. This answer has three significant digits, which is appropriate given the number of significant digits in the original problem.

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