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OpenStudy (anonymous):

what would be the pressure of a given mass of a gas, if its volume and temperature are doubled ............. ?

OpenStudy (anonymous):

Considering an ideal gas, you can use the relation: \[PV=nRT\] taking \(n\) (the nomber of molecules) as a constant, since the mass is not changing, and since \(R\) is the ideal gas constant we get: \[\frac{PV}{T}=constant\] so if you double the volume and temperature you must get the same constant value for that expression, and then: \[\frac{P_1V_1}{T_1}=\frac{P_2(2V_1)}{2T_1}\] what is the relation between the new pressure \(P_2\) and \(P_1\) ??

OpenStudy (anonymous):

he wants complement one word

OpenStudy (anonymous):

consentration

OpenStudy (anonymous):

concentration !

OpenStudy (anonymous):

that word is contained in the response to my question: what is the relation between the new pressure \(P_2\) and \(P_1\) ?? What happens if we cancel out the 2's in the right hand side?

OpenStudy (anonymous):

yes concentration

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