A 1.30-L sample of argon gas is at 1.02 atm and 21.5 degrees Celsius. a. what mass of argon is in the container. b. If the temperature is raised to 500 degrees celsius while the volume remains constant, calculate the new pressure, in stmospheres.

a. To calculate the mass of the argon gas in the container, we need to use the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in Liters)
n = number of moles of gas
R = ideal gas constant (0.08206 L·atm/(mol·K))
T = temperature (in Kelvin)

First, we need to convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 21.5 + 273.15
T(K) = 294.65 K

Now we can rearrange the ideal gas law equation to solve for n (number of moles of gas):

n = PV / RT

Substituting the given values:
n = (1.02 atm) * (1.30 L) / (0.08206 L·atm/(mol·K) * 294.65 K)

Now we can calculate the number of moles:
n = 0.0583 moles

To find the mass of the argon gas, we need to use the molar mass of argon, which is 39.95 g/mol.

Mass = n * molar mass of argon
Mass = 0.0583 moles * 39.95 g/mol

Calculating the mass:
Mass = 2.33 grams

Therefore, the mass of argon in the container is 2.33 grams.

b. If the volume remains constant, the only variable that changes is the temperature. To calculate the new pressure at 500 degrees Celsius, we need to convert it to Kelvin:

T(K) = 500 + 273.15
T(K) = 773.15 K

Since the volume is constant, we can use the Gay-Lussac's law, which states that the pressure of a gas is directly proportional to its temperature:

P1 / T1 = P2 / T2

Substituting the values:
1.02 atm / 294.65 K = P2 / 773.15 K

Solving for P2:
P2 = (1.02 atm * 773.15 K) / 294.65 K

Calculating the new pressure:
P2 = 2.68 atm

Therefore, the new pressure, in atmospheres, is 2.68 atm.

To calculate the mass of argon in the container, we can use the ideal gas law equation:

PV = nRT

Where:
P = Pressure
V = Volume
n = Number of moles
R = Ideal gas constant (0.08206 L·atm/mol·K)
T = Temperature

a. First, let's convert the given temperature from Celsius to Kelvin:
T1 = 21.5 + 273.15 = 294.65 K

The ideal gas law equation can be rearranged to solve for the number of moles (n) of Argon:
n = PV / RT

Plugging in the known values:
n = (1.02 atm) * (1.30 L) / ((0.08206 L·atm/mol·K) * (294.65 K))

Now, we can calculate the mass using the molar mass (atomic mass) of Argon:
Molar mass of Argon = 39.95 g/mol

Mass of Argon = n * molar mass of Argon

b. To calculate the new pressure when the temperature is raised to 500 degrees Celsius while the volume remains constant, we can use Charles's Law, which states that at constant pressure, the volume of a gas is directly proportional to its temperature.

So, we can write:
V1 / T1 = V2 / T2

Given that the volume remains constant, V1 / V2 = 1, and the equation simplifies to:
T1 / T2 = P1 / P2

We can rearrange the equation to solve for the new pressure (P2):
P2 = P1 * T2 / T1

Now, let's calculate the new pressure:
P2 = (1.02 atm) * (500 + 273.15 K) / 294.65 K

With this information, we can now calculate the answers to both parts of the question.

PV=nRT

where
P= pressure in atm
V= volume in L
n =(mass of Ar)/(molar mass of Ar)
R=0.082058 L atm mol-1 K-1
T= temperature in Kelvins