A tank in a room at 19 Degree C is initially open to the atmosphere on a day whe

Question

A tank in a room at 19 Degree C is initially open to the atmosphere on a day when the barometric pressure is 102 kPa. A block of dry ice (solid carbon dioxide) with a mass of 15.7 kg is dropped into the tank, which is then sealed. the reading on the tank pressure gauge initially rises very quickly, then much more slowly, eventually reaching a value of 3.27 MPa. How many kg mols of air were in the tank a. initially? Neglect the volume occupied by Co_2 in the solid state, and assume that a negligible amount of Co_2 escapes prior to sealing of the tank. What is the final density (g/liter) of the gas in the tank?

Explanation / Answer

when the tank was open to the atomosphere it had pressure P(air) = 102 kPa

Let's convert this value to atm

102 kPa * 1 atm/ 101.3 kPa = 1.007 atm

at this point, let's say moles of air = n (air)

Let's say volume of the tank = V

T = 292 K

Therefore, according to ideal gas equation we have

PV = nRT

P(air) V = n (air) * R* 292

V = n (air) *R* 292/P (air)

R = 0.0821 -atm / mol K ( R is gas constant)

Substituting the values, we get

V = n (air) * 0.0821 * 292 / 1.007 ---------------- (equation 1)

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Now let's check what do we have when we add dry ice

moles of dry ice added = 15.7 kg * 1000 g / 1 kg * 1 mol / 44 g = 356.8 mol

Therefore we have n (CO2) = 356.8 mol

Total pressure when the tank was closed P (total) = 3.27 MPa

Let us convert this to atm

3.27 MPa * 10^6 Pa/ 1 MPa * 1atm/ 101325 Pa = 32.27 atm

P (total) = 32.27 atm

total moles when the tank was closed n (total) = n (air) + n (CO2)

Volume and Temperature remains the same

Applying ideal gas law again we get,

P(total) V = n(total) * R * 292

32.27 * V = [ n(air) + 356.8 mol ] * 0.0821 * 292

V = [ n(air) + 356.8 mol ] * 0.0821 * 292 / 32.27 ----------------( equation 2)

Equation 1 and 2both have V on left side, therefore we can equate their right sides

n (air) * 0.0821 * 292 / 1.007 =    [ n(air) + 356.8 mol ] * 0.0821 * 292 / 32.27

On simplification we get

n (air) / 1.007 = n(air) + 356.8 / 32.27 .........( 0.0821 & T (292) get canceled as they are common on both sides)

on cross multiplication,

32.27 *n(air) = 1.007 [ n (air) + 356.8]

32.27n (air) = 1.007 n (air) + 359.3

32.27 n (air) - 1.007 n (air) = 359.3

31.26 n (air) = 359.3

n (air) = 359.3/31.26

n (air) = 11.5 mol

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part b

Using mol value of air, we can find volume of air as follows

P(air) * V (air) = n (air) * R * T

1.007 atm * V (air) = 11.5 mol * 0.0821 L-atm/mol K * 292 K

V (air) = 274 L

Since only air was present in the tank initially, we can assume that volume of te tank is 274 L

Now mass of air in the tank is

density of air * volume

Density of air = 1.225g/L

mass of air = 1.225 g/L * 274 L = 335.7 g

mass of CO2 = 15.7 kg = 15700 g

total mass = 16035.7 g

density of the gas inside the tank = 16035.7 g/ 274 L

density of gas = 58.5 g/L

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