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A steel cylinder contains 150.0 mol argon gas at a temperature of 25°C and a pressure of 7.04 MPa. After some argon has been used, the pressure is 2.00 MPa at a temperature of 19°C. What mass (g) of argon remains in the cylinder?

Respuesta :

Answer:

There remains 1737 grams of argon gas

Explanation:

Step 1: Data given

Moles of argon gas = 150.0 moles

Temperature of the argon gas = 25.0 °C = 298 K

Initial Pressure = 7.04 MPa

Final pressure = 2.00 MPa

Final temperature = 19.0 °C

Step 2: Calculate moles of argon

p*V = n*R*T

We can say:

P/(n*T) = R/V

or P1/(n1*T1) = P2/(n2*T2)

⇒ with P1 = initial pressure = 7.04 MPa = 69.4794 atm

⇒ with n1 = initial number of moles = 150.0 moles

⇒ with T1 = The initial temperature = 25.0 °C = 298 K

⇒ with P2 = the final pressure = 2.00 MPa = 19.7385 atm

⇒ with n2 = the final number of moles = TO BE DETERMINED

⇒ with T2 = the final temperature = 19.0 °C = 292 K

7.04 / (150.0 * 298) = 2.00/(n2 * 292)

1.575 *10^-4 = 2/(292n2)

292n2 = 2/1.575 *10^-4

n2 = 43.49 moles

Step 3: Calculate mass of argon

Mass = moles * molar mass

Mass = 43.49 * 39.95 g/mol

Mass argon = 1737 grams

There remains 1737 grams of argon gas

onyebuchinnaji

The mass of argon remaining in the cylinder is 1,740 g.

The given parameters;

  • initial number of moles of the gas, n₁ = 150
  • initial temperature, T₁ = 25 ⁰C = 298 K
  • initial pressure of the gas, P₁ = 7.04 MPa
  • final pressure of the gas, P₂ = 2 MPa
  • final temperature of the gas, T₂ = 19 ⁰C = 292 k

The final number of moles of the gas is calculated as follows;

[tex]PV = nRT\\\\\frac{P_1}{n_1 T_1 } = \frac{P_2}{n_2 T_2} \\\\n_2 = \frac{n_1 T_1 P_2}{P_1 T_2} \\\\n_2 = \frac{150 \times 298 \times 2}{7.04 \times 292} \\\\n_2 = 43.5 \ moles[/tex]

The molar mass of argon gas is given as;

molar mass = 40 g/mol

The mass of argon remaining in the cylinder is calculated as;

M = n x molar mass

M = 43.5 x 40

M = 1,740 g

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