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What volume is occupied by 24.3 g of argon gas at a pressure of 1.38 atm and a temperature of 414 K ? Express your answer with the appropriate units. Part B Would the volume be different if the sample were 24.3 g of helium (under identical conditions)? The volume would be greater for helium gas. The volume would be the same for helium gas. The volume would be lower for helium gas.

User Dikkini
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Final answer:

The volume occupied by 24.3 g of argon gas at the given conditions is 235.8 L.

Step-by-step explanation:

To solve this problem, we can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to convert the temperature from Celsius to Kelvin by adding 273.15: T = 414 K.

From the equation, we can rearrange it to solve for volume: V = (nRT) / P.

Substituting the given values: V = (24.3 g / 39.95 g/mol) * (0.0821 L*atm/mol*K) * 414 K / 1.38 atm = 235.8 L.

Therefore, the volume occupied by 24.3 g of argon gas at a pressure of 1.38 atm and a temperature of 414 K is 235.8 L.

User Ashbay
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Final answer:

The volume occupied by 24.3 g of argon gas at a pressure of 1.38 atm and a temperature of 414 K is 11.3 L.

Step-by-step explanation:

To find the volume occupied by 24.3 g of argon gas at a pressure of 1.38 atm and a temperature of 414 K, we can use the ideal gas law equation: PV = nRT.

First, we need to determine the number of moles of argon gas. The molar mass of argon (Ar) is 39.95 g/mol, so we divide the mass of the gas (24.3 g) by the molar mass to get the number of moles: 24.3 g ÷ 39.95 g/mol = 0.608 mol.

Next, we plug the values into the ideal gas law equation: (1.38 atm) × V = (0.608 mol) × (0.0821 L·atm/mol·K) × (414 K). Solving for V, we get: V = (0.608 mol × 0.0821 L·atm/mol·K × 414 K) ÷ 1.38 atm = 11.3 L.

Therefore, the volume occupied by the argon gas is 11.3 L.

User Paul Asjes
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