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300 K temperature at 2.0xx10^(4) Pa pressure volume of 0.09 mole CO_(2) gas is 112.0xx10^(-3)m^(3)then calculate volume at 4.0xx10^(4) Pa ? |
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Answer» <P> Solution :`64.2xx10^(-3)m^(3)``p_(1)V_(1)=p_(2)V_(2)` `therefore V_(2)=(p_(1)V_(1))/(p_(2))=(2.0xx10^(4)Pa xx112.0xx10^(-3)m^(3))/(4.0xx10^(4)Pa)` `= 64.2xx10^(-3)m^(3)` (Note : As the pressure divided the volume become half.) |
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