document.write( "Question 189916: i can really use your help!\r
\n" ); document.write( "\n" ); document.write( "If a gas was in a 7.5 liter flexible container under 4 atmospheres of pressure, how would the volume of the gas change and what would it be if the pressure was increased to 6.5 atmospheres, keeping all other variables constant? \r
\n" ); document.write( "\n" ); document.write( "a. The volume would decrease to about 4.6 liters.
\n" ); document.write( "b. The volume would decrease to about 3.5 liters.
\n" ); document.write( "c. The volume would increase to about 12.2 liters.
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Algebra.Com's Answer #142602 by schrammbledeggs(41)\"\" \"About 
You can put this solution on YOUR website!
For this chemistry problem you want to use the equation P1V1 = P2V2 where:\r
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\n" ); document.write( "\n" ); document.write( "P1 = Initial Pressure
\n" ); document.write( "V1 = Initial Volume
\n" ); document.write( "P2 = Changed Pressure
\n" ); document.write( "V2 = Changed Volume\r
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\n" ); document.write( "\n" ); document.write( "We're given values for everything but V2 so let's plug in the other values:\r
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\n" ); document.write( "\n" ); document.write( "4 atm * 7.5 liters = 6.5 atm * V2\r
\n" ); document.write( "\n" ); document.write( "30 atm/liter = 6.5atm * V2\r
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\n" ); document.write( "\n" ); document.write( "Divide both sides by 6.5 atm:\r
\n" ); document.write( "\n" ); document.write( "30 atm/liter / 6.5 atm = V2
\n" ); document.write( "4.615 ~ V2\r
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\n" ); document.write( "\n" ); document.write( "So the answer is a. The volume would decrease to about 4.6 liters.\r
\n" ); document.write( "\n" ); document.write( ":)
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