Calculate the volume occupied by 14-0 g N2 at 200 K and 8-21 atm pressure if PcVcRTc-3-8 and PrVrTt-2-2 -write the value to the nearest integer-

14 g N_2→ 0.5 mole T=200k, P=8.21 atm P_cV_c/RT_c=3/8, P_rV_r/T_r=2.2 P_r=P/P_c, V_r=V/V_c, T_r=T/T_c SO, (P_cP_r)(V_cV_r)/R(T_cT_r)=3/ ...

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Standard X

Chemistry

Mole Concept

Calculate the...

Question

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]

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Solution

$14 g N_{2} \to 0.5 m o l e$
$T = 200 k, P = 8.21 a t m$
$\frac{P_{c} V_{c}}{R T_{c}} = \frac{3}{8}, \frac{P_{r} V_{r}}{T_{r}} = 2.2$
$P_{r} = \frac{P}{P_{c}}, V_{r} = \frac{V}{V_{c}}, T_{r} = T / T_{c}$ SO,
$\frac{(P_{c} P_{r}) (V_{c} V_{r})}{R (T_{c} T_{r})} = \frac{3}{8} \times 2.2 \Rightarrow \frac{P V}{R T} = \frac{3}{2} \times 2.2$
$V = \frac{3}{8} \times \frac{2.2 \times 0.0821 \times 200}{8.21} = 1.65 L$
so volume of 0.5 mole $N_{2}$ (0.5 250 $N_{2}$ 5 5:) $= 1.65 \times 0.5 = 0.825 L$

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Calculate the volume occupied by 14.0 g N2 at 200 K and 8.21 atm pressure if PcVcRTc=38 and PrVrTt=2.2 [write the value to the nearest integer]

14gN2→0.5moleT=200k,P=8.21atmPcVcRTc=38,PrVrTr=2.2Pr=PPc,Vr=VVc,Tr=T/Tc SO,(PcPr)(VcVr)R(TcTr)=38×2.2⇒PVRT=32×2.2V=38×2.2×0.0821×2008.21=1.65Lso volume of 0.5 mole N2 (0.5 250 N2 5 5:)=1.65×0.5=0.825L

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]