Calculate pressure exerted by 1 mole of a van der Waal gas at a temperature of 8-0-0821 K in a 1-2 L container- if the volume of the molecule is assumed to be negligible and van der Waal-s constant a -2 atm L 2 mol -2

Using van der Waal's equation for 1 mol: ( P + aV^2_m ) (V_m b)= RT Since the volume is negligible, neglecting b in the above equation we get: ( P + aV^2_m ) V ...

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Standard XII

Chemistry

Van der Waal's Forces

Calculate pre...

Question

Calculate pressure exerted by 1 mole of a van der Waal gas at a temperature of $\frac{8}{0.0821} K$ in a $\frac{1}{2} L$ container, if the volume of the molecule is assumed to be negligible and van der Waal's constant $a = 2 {atm L}^{2} {mol}^{- 2}$

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Solution

Using van der Waal's equation for 1 mol:
$(P + \frac{a}{V_{m}^{2}}) (V_{m} - b) = R T$
Since the volume is negligible, neglecting $b$ in the above equation we get:
$(P + \frac{a}{V_{m}^{2}}) V_{m} = R T$
Given : $T = \frac{8}{0.0821} K$
$P V_{m} + \frac{a}{V_{m}} = R \times \frac{8}{0.0821}$
$P V_{m}^{2} - 8 V_{m} + a = 0$
According to the question, $V_{m} = \frac{1}{2} L = 0.5 L$ so,
$P (0.5)^{2} - 8 (0.5) + 2 = 0$
$P = 8 atm$

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Calculate pressure exerted by 1 mole of a van der Waal gas at a temperature of 80.0821K in a 12 L container, if the volume of the molecule is assumed to be negligible and van der Waal's constant a=2 atm L2 mol−2

Using van der Waal's equation for 1 mol: (P+aV2m)(Vm−b)=RT Since the volume is negligible, neglecting b in the above equation we get: (P+aV2m)Vm=RT Given : T=80.0821K PVm+aVm=R×80.0821 PV2m−8Vm+a=0 According to the question, Vm=12 L=0.5 L so, P(0.5)2−8(0.5)+2=0 P=8 atm

Calculate pressure exerted by 1 mole of a van der Waal gas at a temperature of $\frac{8}{0.0821} K$ in a $\frac{1}{2} L$ container, if the volume of the molecule is assumed to be negligible and van der Waal's constant $a = 2 {atm L}^{2} {mol}^{- 2}$