Question

An air bottle used for scuba diving breathing apparatus has a volume of 30 l. After it has been used, the pressure and temperature of the remaining air in the bottle are 0.3 MPa and 24 ◦C. Before the apparatus is used again, it is charged with air at temperature of 50 ◦C until the pressure in the bottle reaches 6 MPa. The charging processes is fast, and can be assumed to be adiabatic.

Questions: (a) Calculate the mass of air to be added to the bottle.

(b) What is the temperature of the air in the bottle at the end of the charging process?

Answer 1

#(b):

After use:

P₁ = 0.3 MPa

T₁ = 24+273 = 297 K

After charged:

P₂ = 6 MPa

T₂ = ??

V = Constant = 30 L

specific heat ratio of air, k = 7/5 or 1.4

Since the process is adiabatic and V is constant,

P^1-k * T^k = constant

=> P₁^1-k * T₁^k = P₂^1-k * T₂^k

=> (0.3)^1-1.4 * (297)^1.4 = (6)^1-1.4 * (T₂)^1.4

=> (T₂)^1.4 = (0.3)^-0.4 * (297)^1.4 / (6)^-0.4 = 9600

=> T₂ = 699 K or 426 ^oC (Answer)

Hence temperature at the end of charging process is 699 K or 426 ^oC.

#(a):

Moles of air after use:

P = 0.3 MPa * (1 atm / 0.101325 MPa) = 2.96 atm

T = 24+273 = 297 K

V = 30 L

=> Moles of air inside the empty bottle after use, n = PV/RT = (2.96 atm * 30 L) / (0.08206 L.atm.mol^-1.K^-1*297 K)

= 3.6445 mol

Moles after charging:

P = 6 MPa * (1 atm / 0.101325 MPa) = 59.2154 atm

T = 699 K

V = 30 L

=> Moles of air inside the empty bottle after charging, n = PV/RT = (59.2154 atm * 30 L) / (0.08206 L.atm.mol^-1.K^-1*699 K)

= 30.97 mol

=> Moles of air added = (Moles after charging) - (moles after use)

= 30.97 mol - 3.6445 mol

= 27.3255 mol

=> mass of air added = 27.3255 mol * 28.97 g/mol = 791.6 g (Answer)