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kiruha [24]
2 years ago
11

Cylinders of compressed gas are typically filled to a pressure of 200 bar. For oxygen, what would be the molar volume at this pr

essure and 25 °C based on (i) the perfect gas equation, (ii) the van der Waals equation? For oxygen, a = 1.364 dm6 atm mol−2, b = 3.19 × 10−2 dm3 mol−1.
Chemistry
1 answer:
mars1129 [50]2 years ago
7 0

Answer:

a

 V  =   0.124 \ Liters

b

V  =  0.112 \  Liters

Explanation:

From the question we are told that

  The pressure of compressed gas is P  =  200 \  bar = \frac{200}{1.013}=  197.4 \ atm

  The  temperature is  T  =  25^oC =  25 + 273 = 298 \  K

Generally the perfect gas equation is mathematically represented as

     PV =  nRT

substituting 0.08206 L-atm/mol-K  for R and  1 mole for n

We have that

     V  =  \frac{1 *  0.08206 * 298 }{ 197.4}

       V  =   0.124 \ Liters

Generally the van der Waals equation is mathematically represented as

       nRT  =  [P + \frac{n^2 * a }{V^2 } ][V - nb]

=>1 *  0.08206 *298  =  [197.4 + \frac{1^2 *  1.364}{V^2 } ][V - 1 * 3.19 * 10^{-2}]

=>    V  =  0.112 \  Liters

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Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(74.0 days) = 9.365 x 10⁻³ day⁻¹.

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t is the time of the reaction (t = ??? day).

a is the initial concentration of Ir-192 (a = 560.0 dpm).

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(9.365 x 10⁻³ day⁻¹)(t) = ln(560.0 dpm)/(35.0 dpm).

(9.365 x 10⁻³ day⁻¹)(t) = 2.773.

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Check the explanation

Explanation:

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= 8.00*10-3 mol.

Mols Sr(OH)2 corresponding to 10.0 mL of 0.100 M solution =

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Sr(OH)2 (aq) ----------> Sr2+ (aq) + 2 OH- (aq)

As per the stoichiometric equation,

1 mol Sr(OH)2 = 2 mols OH-.

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HC4H7O2 (aq) + OH- (aq) ------------> H2O (l) + C4H7O2- (aq)

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Change (mol)      -2.00*10-3       -2.00*10-3                           -        +2.00*10-3

After (mol)           6.00*10-3                0                                  -          2.00*10-3

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The initial concentrations are obtained by dividing the numbers of moles by the volume, 0.05 L.

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Change (M)        -0.0400            -0.0400                            -           +0.0400

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