SOLUTION: The volume V of an ideal gas varies directly with the temperature T and inversely with the pressure P. A cylinder contains oxygen at a temperature of 310 degrees K and a pressure o

Algebra ->  Rational-functions -> SOLUTION: The volume V of an ideal gas varies directly with the temperature T and inversely with the pressure P. A cylinder contains oxygen at a temperature of 310 degrees K and a pressure o      Log On


   



Question 1137495: The volume V of an ideal gas varies directly with the temperature T and inversely with the pressure P. A cylinder contains oxygen at a temperature of 310 degrees K and a pressure of 18 atmospheres in a volume of 120 liters. What is the pressure if the volume is decreased to 100 liters and the temperature is increased to 320 degrees K?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
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The volume V of an ideal gas varies directly with the temperature T and inversely with the pressure P.
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V=kT%2FP-------using lower-case k as the variation constant


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cylinder contains oxygen at a temperature of 310 degrees K and a pressure of 18 atmospheres in a volume of 120 liters.
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120=k%2A310%2F18

k=120%2A18%2F310

k=12%2A18%2F31

k=216%2F31
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highlight%28V=%28216%2F31%29%28T%2FP%29%29


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What is the pressure if the volume is decreased to 100 liters and the temperature is increased to 320 degrees K?
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This questions asks for P, given certain values of V and T.

Answer by ikleyn(52814) About Me  (Show Source):
You can put this solution on YOUR website!
.
From the condition, you have this general formula

    V = k%2A%28T%2FP%29,

where " k " is  a constant value.
    


From the condition, you also have these two expressions for two states of the gas:

    120 = k%2A%28310%2F18%29    (1)

and

    100 = k%2A%28320%2FP%29,    (2)

where P is the unknown pressure under the question.



Divide expression (1) by expression (2) (both sides).  You will get


    120%2F100 = %28%28310%2F18%29%29%2F%28%28320%2FP%29%29,   or

     1.2 = %28310%2AP%29%2F%2818%2A320%29.

which implies

     P = %281.2%2A18%2A320%29%2F310 = 22.297   atmospheres.    ANSWER

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It is the shortest way to solve such problems.

The other approach with calculating the constant value " k " forces you to make unnecessary calculations on the way to the answer.