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# 1 mole of monoatomic gas confined in a 22.5 litre flask at 273 K exerts a pressure of 0.98 atm whereas expected pressure is 1 atm. Has the gas behaved ideally? determine a, b and boyle's temperature.

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Solution

## Dear Student, At temperature = 273 K, 1 mole of gas should exert a pressure of 1 atm and volume of 22.4 L. But it is exerting a pressure of 0.98 atm and has volume of 22.5 L. So, the gas is not behaving ideally. For real gases of 1 mole, $\left(P+\frac{a}{{V}^{2}}\right)\left(V-b\right)=RT$ $\left(P+\frac{a}{{V}^{2}}\right)=1.0\phantom{\rule{0ex}{0ex}}\left(0.98+\frac{a}{{V}^{2}}\right)=1.0\phantom{\rule{0ex}{0ex}}\frac{a}{{V}^{2}}=0.02\phantom{\rule{0ex}{0ex}}\frac{a}{22.{5}^{2}}=0.02$ a = 10.125 V-b = 22.4 22.5-b = 22.4 b = 0.1 Boyles temperature = Tb = $\frac{a}{Rb}=\frac{10.125}{0.082×0.1}$= 1234 K Boyle's temperature = 1234 K Regards  Suggest Corrections  0      Similar questions
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