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(A) According to the ideal gas equation,$PV = nRT$,where $P$ is pressure,$V$ is volume,$n$ is the number of moles,$R$ is the universal gas constant,an

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$H_2$

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(A) According to the ideal gas equation,$PV = nRT$,where $P$ is pressure,$V$ is volume,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is temperature. Since $V$,$n$,and $T$ are constant for all four gases,the pressure $P$ is given by $P = \frac{nRT}{V}$. Since $n$,$R$,$T$,and $V$ are identical for all gases,the pressure exerted by each gas will be the same. However,in real gas behavior,the pressure exerted depends on the compressibility factor and intermolecular forces. For ideal gases,the pressure is equal. If the question implies real gases,the pressure exerted by a gas is influenced by the van der Waals equation: $(P + \frac{an^2}{V^2})(V - nb) = nRT$. For gases with smaller molecular size and weaker intermolecular forces (like $H_2$),the deviation from ideal behavior is minimal,and they exert pressure closest to the ideal value. Given the options and the standard context of such problems,if all are treated as ideal,the pressure is the same. If the question implies which gas behaves most ideally or exerts the highest pressure due to lower attractive forces,$H_2$ is the correct choice.

Four vessels of same volume consist of equal moles of four gases $H_2$,$Cl_2$,$N_2$,and $O_2$ separately at the same temperature. The pressure exerted by the gas is maximum for -

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