Question

What volume is occupied by 26.6 g of argon gas at a pressure of 1.29 atm and a temperature of 355 K ?

Answer 1

The volume occupied by 26.6 g of argon gas at a pressure of 1.29 atm and a temperature of 355 K is 15.91 L.

Step-by-step explanation:

The question is asking for the volume occupied by 26.6 g of argon gas at a pressure of 1.29 atm and a temperature of 355 K. We can use the ideal gas law to solve this problem. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

We need to convert the given mass of argon gas to moles. The molar mass of argon is about 39.948 g/mol. Using the formula: moles = mass / molar mass, we can find that moles = 26.6 g / 39.948 g/mol = 0.667 mol.

Now, we can rearrange the ideal gas law to solve for V:

V = (nRT) / P

Plugging in the known values, we get: V = (0.667 mol)(0.0821 L/mol·K)(355 K) / 1.29 atm = 15.91 L.

Answer 2

V = 15 L

Step-by-step explanation:

PV = nRT

P = 1.29 atm, T= 355 K mass argon 40g/mol

n = mass/ molecular mass; n = 26.6 g/40 g/mol; n = 0.665 mol

V = (0,665 mol x 0.082 x 355 K) / 1.29 atm = 15 L