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3 moles of a mono-atomic gas (gamma=5//3) is mixed with 1 mole of a diatomic gas (gamma=7//3). The value of gamma for the mixture will be |
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Answer» `9//11` Here, `n_(1)=3,gamma_(1)=(5)/(3),n_(2)=1,gamma_(2)=(7)/(3)` `therefore gamma_("mixture")=(=(3xx5//3)/([5//3-1])+(1xx7//3)/([7//3-1]))/((3)/([5//3-1])+(1)/([7//3-1]))=((15)/(2)+(7)/(4))/((9)/(2)+(3)/(4))=(37)/(21)` Note : If we TAKE `gamma_(2)=7//5` `gamma_("mixture")=(=(3xx5//3)/([5//3-1])+(1xx7//5)/([7//5-1]))/((3)/([5//3-1])+(1)/([7//5-1]))=((15)/(2)+(7)/(2))/((9)/(2)+(5)/(2))=(22)/(14)=(11)/(7)` |
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