Question

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?

Answer 1

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

Answer 2

Answer: 1739.6g

Explanation:

P1 = 7.04MPa

T1 = 25°C = 25 + 273 = 298K

n1 = 150mol

P2 = 2MPa

T2 = 19°C = 19 + 273 = 292K

n2 =?

n1T1/P1 = n2T2/P2

(150x298)/7.04 = (n2x292)/2

7.04xn2x292 = 150x298x2

n2 = (150x298x2) /(7.04x292)

n2 = 43.49moles

Molar Mass of Ar = 40g/mol

Mass = n x molar Mass = 43.49x40 = 1739.6g