If pressure of CO 2 real gas in a container is given by P - RT -2 V - b - a -4 V 2 then mass of the gas in container isA- 11 gmB- 22 gmC- 33 gmD- 44 gm

The correct option is B 22 gmVander wal's gas equation for μ mole of real gas (P + μ^2α/V^2)(V μb) = μRT ⇒ P = μ RT/V μ b μ^2/V^2 on comparing the given ...

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Standard X

Chemistry

Gay Lussac's Law

If pressure o...

Question

If pressure of $C O_{2}$ (real gas) in a container is given by P = $\frac{R T}{2 V - b} - \frac{a}{4 V^{2}}$ then mass of the gas in container is

11 gm

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22 gm

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33 gm

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44 gm

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Solution

The correct option is B 22 gm
Vander wal's gas equation for $μ$ mole of real gas
(P + $\frac{μ^{2} α}{V^{2}}$ )(V - $μ$ b) = $μ$ RT $\Rightarrow$ P = $\frac{μ R T}{V - μ b} - \frac{μ^{2}}{V^{2}}$
on comparing the given equation with this standard equation we get $μ = \frac{1}{2}$ . Hence $μ$ = $\frac{m}{M} \Rightarrow$ mass of gas
m = $μ M$ = $\frac{1}{2} \times 44$ = 22gm

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If pressure of CO2 (real gas) in a container is given by P = RT2V−b−a4V2 then mass of the gas in container is

The correct option is B 22 gmVander wal's gas equation for μ mole of real gas (P + μ2αV2)(V - μb) = μRT ⇒ P = μRTV−μb−μ2V2 on comparing the given equation with this standard equation we get μ=12. Hence μ = mM⇒ mass of gas m = μM = 12×44 = 22gm

If pressure of $C O_{2}$ (real gas) in a container is given by P = $\frac{R T}{2 V - b} - \frac{a}{4 V^{2}}$ then mass of the gas in container is