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