Calculate the volume occupied by 16 gram O 2 at 300 K and 8-31 Mpa if P c V - RT c -3-8 and P r V r - T r -2-21 - R -8-314 Mpa - K - mol A- 124-31 litB- 248-62 LC- 124-31 mLD- 248-62 mL

The correct option is C 124.31 mL 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 = 124.3 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

124.31 mL

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

248.62 mL

No worries! We‘ve got your back. Try BYJU‘S free classes today!

248.62 L

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
124.31 mL

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

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

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

Suggest Corrections

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]

The volume occupied by 2 mol of $N_{2}$ at 200 K and 10.1325 MPa pressure is

(Given that $\frac{P_{c} V_{C}}{R T_{C}} = \frac{3}{8}$ and $\frac{P_{r} V_{r}}{T_{r}} = 2.21$ ), where $P_{r}, V_{r}$ and $T_{r}$ are reduced pressure, reduced volume and reduced temperature.

Join BYJU'S Learning Program

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

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

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

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