A commercial cylinder contains 6-01 m3 of O2 at 15-18 M Pa and 21-C- The critical constants for O2 are Tc-118-4-C- Pc-50-1 atm- Determine the reduced pressure and reduced temperature for O2 under these conditions-

The correct option is A Pr=2.99, Tr=1.90 The reduced pressure π = P/P_c = #160; #160; 15.18 MPa/50.1 × 0.1 MPa = 2.99 The reduced temperature ...

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Standard XII

Chemistry

Surface Tension

A commercial ...

Question

A commercial cylinder contains 6.01 $m^{3}$ of $O_{2}$ at 15.18 M Pa and $21^{\circ} C$ . The critical constants for $O_{2}$ are $T_{c} = - {118.4}^{\circ} C, P_{c} = 50.1$ atm. Determine the reduced pressure and reduced temperature for $O_{2}$ under these conditions.

P r = 2.99, T r = 1.90

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P r = 1.9, T r = 2.99

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P r = 5.98, T r = 3.80

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None of these

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Solution

The correct option is A $P r = 2.99, T r = 1.90$
The reduced pressure $π = \frac{P}{P_{c}} = \frac{15.18 M P a}{50.1 \times 0.1 M P a} = 2.99$

The reduced temperature $θ = \frac{T}{T_{c}} = \frac{21 + 273}{273 - 118.4} = \frac{294}{154.6} = 1.90$

Hence, the correct option is $A$

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and he are given below The gas which has smallest value of van der Waal's constant 'b' is:

Name of the gas	$P_{c}$ $(N m^{- 2})$	$T_{c}$ (K)
$O_{2}$	$5 \times 10^{\circ}$	153
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A commercial cylinder contains 6.01 m3 of O2 at 15.18 M Pa and 21∘C. The critical constants for O2 are Tc=−118.4∘C,Pc=50.1 atm. Determine the reduced pressure and reduced temperature for O2 under these conditions.

The correct option is A Pr=2.99,Tr=1.90The reduced pressure π=PPc=15.18MPa50.1×0.1MPa=2.99The reduced temperature θ=TTc=21+273273−118.4=294154.6=1.90Hence, the correct option is A

A commercial cylinder contains 6.01 $m^{3}$ of $O_{2}$ at 15.18 M Pa and $21^{\circ} C$ . The critical constants for $O_{2}$ are $T_{c} = - {118.4}^{\circ} C, P_{c} = 50.1$ atm. Determine the reduced pressure and reduced temperature for $O_{2}$ under these conditions.

The correct option is A $P r = 2.99, T r = 1.90$
The reduced pressure $π = \frac{P}{P_{c}} = \frac{15.18 M P a}{50.1 \times 0.1 M P a} = 2.99$

The reduced temperature $θ = \frac{T}{T_{c}} = \frac{21 + 273}{273 - 118.4} = \frac{294}{154.6} = 1.90$

Hence, the correct option is $A$