The volume occupied by 2 mol of N 2 at 200 K and 10-1325 MPa pressure is Given that P c V C - RT C -3-8 and Pr r V r - T r -2-21 - where P r - V r and T r are reduced pressure- reduced volume and reduced temperature-A- 544 cm 3B- 272-0 cm 3- 20-7 cm 3D- 136 cm 3

The correct option is B 272.0 cm^3 P_cV_c/RT_c×P_rV_r/T_r = 3/8× 2.21 or V_m = ( 3 × 2.2/8 ) RT/P V_m = 136.0 cm^3 Required volume = V_m × no. of moles ∴ V ...

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Standard XII

Mathematics

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The volume oc...

Question

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.

$20.7 c m^{3}$

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$544 c m^{3}$

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$272.0 c m^{3}$

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$136 c m^{3}$

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Solution

The correct option is C
$272.0 c m^{3}$

$\frac{P_{c} V_{c}}{R T_{c}} \times \frac{P_{r} V_{r}}{T_{r}} = \frac{3}{8} \times 2.21 o r V_{m} = (\frac{3 \times 2.2}{8}) \frac{R T}{P}$

$V_{m} = 136.0 c m^{3}$

Required volume = $V_{m} \times n o . o f m o l e s$

$∴ V_{N_{2}} = 272 c m^{3}$

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Similar questions

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The volume occupied by 2 mol of N2 at 200 K and 10.1325 MPa pressure is (Given that PcVCRTC=38 and PrVrTr=2.21 ), where Pr,Vr and Tr are reduced pressure, reduced volume and reduced temperature.

The correct option is C 272.0cm3 PcVcRTc×PrVrTr=38×2.21 or Vm=(3×2.28)RTP Vm=136.0cm3 Required volume = Vm×no.ofmoles ∴VN2=272cm3

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.

The correct option is C
$272.0 c m^{3}$

$\frac{P_{c} V_{c}}{R T_{c}} \times \frac{P_{r} V_{r}}{T_{r}} = \frac{3}{8} \times 2.21 o r V_{m} = (\frac{3 \times 2.2}{8}) \frac{R T}{P}$

$V_{m} = 136.0 c m^{3}$

Required volume = $V_{m} \times n o . o f m o l e s$

$∴ V_{N_{2}} = 272 c m^{3}$