A cylinder of argon gas contains 125 mol at a pressure of 150 atm and a temperature of 27°C. After some of the

argon has been used, the pressure is 136 atm at 27°C. What mass of argon remains in the cylinder?
A) 4.23 × 103 g B) 113 g C) 5.52 × 103 g D) 4.52 × 103 g E) 5.00 × 103 g

Well, let's calculate the mass of argon that remains in the cylinder. First, we need to use the ideal gas law equation: PV = nRT

We have the initial pressure (P1) = 150 atm, the final pressure (P2) = 136 atm, the initial number of moles (n1) = 125 mol, and the final number of moles (n2) is what we're trying to find.

Now, let's rearrange the equation to solve for n2:
n2 = (P2/P1) * (n1 * V1) / (R * T)

Since the volume (V1) and temperature (T) remain constant, we can simplify the equation further:
n2 = (P2/P1) * n1

Plugging in the values, we get:
n2 = (136/150) * 125
n2 = 113

Now, we know the number of moles of argon remaining in the cylinder is 113 mol. To find the mass, we'll multiply the number of moles by the molar mass of argon.

The molar mass of argon is approximately 39.95 g/mol.

So, the mass of argon remaining in the cylinder is:
mass = n2 * molar mass
mass = 113 mol * 39.95 g/mol
mass ≈ 4523.35 g

Therefore, the answer is D) 4.52 × 10^3 g.

To solve this problem, we can use the ideal gas law equation, which states:

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, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15

T(K) = 27 + 273.15
T(K) = 300.15 K

Next, let's calculate the initial volume of the cylinder. Since the volume is not given, we can assume it to be constant.

PV = nRT

V = (nRT) / P

V = (125 mol * 0.0821 atm L / mol K * 300.15 K) / 150 atm

V = 125 * 0.0821 * 300.15 / 150
V = 124.137 L

Now, let's calculate the final number of moles by rearranging the equation:

n = PV / RT

n = (136 atm * 124.137 L) / (0.0821 atm L / mol K * 300.15 K)

n = 136 * 124.137 / (0.0821 * 300.15)
n ≈ 67.003 mol

The change in the number of moles is given by:

Δn = initial number of moles - final number of moles
Δn = 125 mol - 67.003 mol
Δn ≈ 57.997 mol

Finally, we can calculate the mass of argon remaining:

Mass = number of moles * molar mass

We need to know the molar mass of argon, which is approximately 39.95 g/mol.

Mass = 57.997 mol * 39.95 g/mol

Mass ≈ 2,317 g

Therefore, the correct answer is not given in the options provided.

To find the mass of argon remaining in the cylinder, we can first calculate the initial number of moles of argon using the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in K)

In this case, we are given:
P1 = 150 atm
n1 = 125 mol
T1 = 27°C = 300 K

We can rearrange the equation to solve for V:

V1 = n1RT1 / P1

Now, we can calculate the initial volume (V1).

V1 = (125 mol)(0.0821 L·atm/mol·K)(300 K) / 150 atm
V1 = 165.15 L

Next, we can calculate the final volume (V2) using the same equation, with the given pressure and temperature:

P2 = 136 atm
T2 = 27°C = 300 K

V2 = (n2)(0.0821 L·atm/mol·K)(300 K) / 136 atm
V2 = 180.14 L

Since the volume of the cylinder remains constant, the initial and final volumes are the same, and we can equate them:

V1 = V2

(125 mol)(0.0821 L·atm/mol·K)(300 K) / 150 atm = n2(0.0821 L·atm/mol·K)(300 K) / 136 atm

125 / 150 = n2 / 136

n2 = (125 mol)(136) / 150
n2 = 113.07 mol

Now, we can calculate the mass of argon remaining using the molar mass of argon (39.95 g/mol):

Mass = n2 * molar mass
Mass = 113.07 mol * 39.95 g/mol
Mass = 4521.37 g

Rounding that to the appropriate number of significant figures, the mass of argon remaining in the cylinder is 4.52 x 10^3 g.

Therefore, the correct answer is D) 4.52 × 10^3 g.

Since the temperature is constant, the number of moles is proportional to the pressure:

P1/P2 = n1/n2
P1 = 150 atm, P2 = 136 atm,
n1 = 125 mol, n2 = _____?
Substitute into the above equation and solve for n2 to get the moles of Ar.
Convert the moles of Ar to grams (1 mole Ar = 39.95g)