The relation between pressure exerted byan ideal gas Pideal and observed pressure Preal is given by the equation- P ideal - P real - an 2- V 2i If pressure is taken in N m -2- number of moles in mol and volume in m 3-calculate the unit of -a-ii What will be unit of -a- when pressure is in atmosphere and volume indm 3 -

The correct option is Given that P_ideal = P_real+an^2/V^2 (i) a=PV^2/n^2 If units of P = Nm^ 2, units of V=m^3, units of n=mol then, units of a=Nm^ ...

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Byju's Answer

Standard XII

Chemistry

Real Gases

The relation ...

Question

The relation between pressure exerted byan ideal gas $(p_{i d e a l})$ and observed pressure $(p_{r e a l})$ is given by the equation,

$P_{i d e a l} = P_{r e a l} + \frac{a n^{2}}{V^{2}}$

(i) If pressure is taken in $N m^{- 2}$ , number of moles in mol and volume in $m^{3}$ ,calculate the unit of 'a'.

(ii) What will be unit of 'a' when pressure is in atmosphere and volume in $d m^{3}$ ?

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Solution

Given that $P_{i d e a l} = P_{r e a l} + \frac{a n^{2}}{V^{2}} (i) a = \frac{P V^{2}}{n^{2}} I f u n i t s o f P = N m^{- 2}, u n i t s o f V = m^{3}, u n i t s o f n = m o l t h e n, u n i t s o f a = \frac{N m^{- 2} (m^{3})^{2}}{(m o l)^{2}} = N m^{4} m o l^{-} 2$

(ii) If units of P = atm, units of V= $d m^{3}$ , units of n = mol

then, units of $a = \frac{P V^{2}}{n^{2}} = \frac{a t m . (d m^{3})^{2}}{(m o l)^{2}} = a t m d m^{6} m o l^{- 2}$

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Given that Pideal = Preal+an2V2(i)a=PV2n2If units of P = Nm−2,units of V=m3,units of n=molthen, units of a=Nm−2(m3)2(mol)2=Nm4mol−2 (ii) If units of P = atm, units of V=dm3 , units of n = mol then, units of a = PV2n2=atm.(dm3)2(mol)2=atm dm6mol−2