Calculate the volume occupied by 16 gram O2 at 300 K and 8-31 Mpa if PcV-RTc - 3-8 and PrVr-Tr - 2-21 - R - 8-314 Mpa-K-mol

PV_m/RT = 3/8× 2.21 ; V_m = 3/8× 2.21 ×RT/P V_m = 3/8× 2.21 ×8.314 × 300/8.314 = 248.625 mL V_O_2 = 16/32× 248.625 = 124.31 mL ...

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Standard XII

Chemistry

Liquefaction of Gases and Critical Constants

Calculate the...

Question

Calculate the volume occupied by 16 gram $O_{2}$ at 300 K and 8.31 Mpa if $\frac{P_{c} V}{R T_{c}} = \frac{3}{8} a n d \frac{P_{r} V_{r}}{T_{r}} = 2.21 ⟮ : R = 8.314 \frac{M p a}{K - m o l} ⟯$

124.31 lit

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124.31 mL

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248.62 mL

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248.62 L

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Solution

The correct option is B
124.31 mL

$\frac{P V_{m}}{R T} = \frac{3}{8} \times 2.21; V_{m} = ⟮ \frac{3}{8} \times 2.21 ⟯ \times \frac{R T}{P}$

$V_{m} = \frac{3}{8} \times 2.21 \times \frac{8.314 \times 300}{8.314} = 248.625 m L$

$V_{O_{2}} = \frac{16}{32} \times 248.625 = 124.31 m L$

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Similar questions

Calculate the volume occupied by 16 gram $O_{2}$ at 300 K and 8.31 Mpa if $\frac{P_{c} V}{R T_{c}} = \frac{3}{8} a n d \frac{P_{r} V_{r}}{T_{r}} = 2.21 ⟮ : R = 8.314 \frac{M p a}{K - m o l} ⟯$

Q. Calculate the volume occupied by 2.0 mole of

N_{2}

at 200 K and 8 atm pressure if

\frac{P_{c} V_{c}}{R T_{c}} = \frac{3}{8}

and

\frac{P_{r} V_{r}}{T_{r}} = 2.4

Q. Calculate the volume occupied by

2.0 m o l e

N_{2}

200 K

and

8.21 a t m

pressure, if

\frac{P_{C} V_{C}}{R T_{C}} = \frac{3}{8}

and

\frac{P_{r} V_{r}}{T_{r}} = 2.4

Q. Calculate the volume occupied by 14.0 g

N_{2}

at 200 K and 8.21 atm pressure if

\frac{P_{c} V_{c}}{R T_{c}} = \frac{3}{8}

and

\frac{P_{r} V_{r}}{T_{t}} = 2.2

[write the value to the nearest integer]

Q. The compressibility factor for

N_{2}

223 K

and

81.06 M P a

1.95

, and at

373 K

and

20.265 M P a

it is

1.10

. A certain mass of

N_{2}

occupies a volume of

1.0 d m^{3}

223 K

and

81.06 M P a

. What is the volume occupied by the same quantity of

N_{2}

373 K

and

20.265 M P a

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Calculate the volume occupied by 16 gram O2 at 300 K and 8.31 Mpa if PcVRTc=38 and PrVrTr=2.21 ⟮:R=8.314 MpaK−mol⟯

The correct option is B 124.31 mL PVmRT=38×2.21;Vm=⟮38×2.21⟯×RTP Vm=38×2.21×8.314×3008.314= 248.625 mL VO2=1632×248.625=124.31 mL

Calculate the volume occupied by 16 gram $O_{2}$ at 300 K and 8.31 Mpa if $\frac{P_{c} V}{R T_{c}} = \frac{3}{8} a n d \frac{P_{r} V_{r}}{T_{r}} = 2.21 ⟮ : R = 8.314 \frac{M p a}{K - m o l} ⟯$

The correct option is B
124.31 mL

$\frac{P V_{m}}{R T} = \frac{3}{8} \times 2.21; V_{m} = ⟮ \frac{3}{8} \times 2.21 ⟯ \times \frac{R T}{P}$

$V_{m} = \frac{3}{8} \times 2.21 \times \frac{8.314 \times 300}{8.314} = 248.625 m L$

$V_{O_{2}} = \frac{16}{32} \times 248.625 = 124.31 m L$