Question

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.

Answer 1

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.

Answer 2

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.