Boyle's Law Calculations help?

I need Help on this problem.

A sample of gas occupies 2.50 L at 1.15 atm of pressure. What is the Volume at standard atmospheric pressure?

My teacher Game me the equation (P1)(V1)=(P2)(V2) and PV=K

I don't really know how to do the problem, setup ect. Can someone help me?

2 Answers

  • 10 years ago
    Favorite Answer

    Boyle's law states that the volume (v) of a gas and its pressure (p) are inversely proportional.

    V= Constant/P

    If we multiply both sides by P, we get: PV = Constant

    This relationship is true because if the pressure increases, the volume decreases, but the product P x V is always equal to the same constant. For two different sets of condtions, we can say that

    P1V1 = Constant = P2V2 or P1V1 = P2V2.

    Where P1 and V1 are the initial pressure and volume of the gas, and P2 and V2 are the final volume and pressure.

    Your problem is given: 2.50 L of a sample gas

    1.15 atm of Pressure

    Find: the volume of the atomosheric pressure

    I am not sure if you have a third variable, but if not then you would use the formula of:

    V(volume) = Constant divided by P (pressure)

    Therefore, 2.50 L is your constant and 1.15 atm is the pressure.

    2.50 L / 1.15 atm = 2.173913043 The average answer would be 2.17 using three significant figures for your answer.

  • ?
    Lv 4
    4 years ago

    Its asking you exa6524b7f26958c1604398aa6ea09224tly what the equation 6524b7f26958c1604398aa6ea09224eans. you start up out with 6524b7f26958c1604398aa6ea092246524b7f26958c1604398aa6ea092246524b7f26958c1604398aa6ea092246524b7f26958c1604398aa6ea092246524b7f26958c1604398aa6ea092246524b7f26958c1604398aa6ea09224 of oxygen at a stress of 9.80 kPa. and you desire to be sure what the hot volu6524b7f26958c1604398aa6ea09224e would be while the stress be6524b7f26958c1604398aa6ea09224o6524b7f26958c1604398aa6ea09224es 9.6524b7f26958c1604398aa6ea092240kPa. so which you plug it in like P1V1=P6524b7f26958c1604398aa6ea09224V6524b7f26958c1604398aa6ea09224 9.80kPa(6524b7f26958c1604398aa6ea092246524b7f26958c1604398aa6ea092246524b7f26958c1604398aa6ea092246524b7f26958c1604398aa6ea092246524b7f26958c1604398aa6ea092246524b7f26958c1604398aa6ea09224)=9.6524b7f26958c1604398aa6ea092240kPaV6524b7f26958c1604398aa6ea09224. this is easiest to isolate the variable you're attempting to discern for (V6524b7f26958c1604398aa6ea09224) by capacity of dividing the two aspects by capacity of P6524b7f26958c1604398aa6ea09224. So next you're able to write V6524b7f26958c1604398aa6ea09224=(9.80kPa[6524b7f26958c1604398aa6ea092246524b7f26958c1604398aa6ea092246524b7f26958c1604398aa6ea092246524b7f26958c1604398aa6ea092246524b7f26958c1604398aa6ea092246524b7f26958c1604398aa6ea09224])/9.6524b7f26958c1604398aa6ea092240kPa in basic terms 6524b7f26958c1604398aa6ea09224ultiply P1 and V1, then divide your answer by capacity of P6524b7f26958c1604398aa6ea09224. while gas 6524b7f26958c1604398aa6ea09224ole6524b7f26958c1604398aa6ea09224ules are uncovered to distinctive pressures, their volu6524b7f26958c1604398aa6ea09224e 6524b7f26958c1604398aa6ea09224hanges a6524b7f26958c1604398aa6ea092246524b7f26958c1604398aa6ea09224ordingly. in basic terms like in the event that they are placed right into a much better 6524b7f26958c1604398aa6ea09224ontainer and the volu6524b7f26958c1604398aa6ea09224e in6524b7f26958c1604398aa6ea09224reases, the stress will de6524b7f26958c1604398aa6ea09224rease. in case you have 3 of the variables, you 6524b7f26958c1604398aa6ea09224an remedy for the fourth in basic terms like quite a few different 6524b7f26958c1604398aa6ea09224athe6524b7f26958c1604398aa6ea09224ati6524b7f26958c1604398aa6ea09224al equation.

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