document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1173943>Question 1173943</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The pressure (P) of a gas varies inversely as the volume (V). What will happen to the volume if the pressure is doubled and all other things remain constant? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #799276 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13206)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=799276\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "The formal definition of inverse variation tells us that the pressure and volume are related by the function<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=P+=+k%2FV ALIGN=MIDDLE  ALT=\"P+=+k%2FV\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "However, computations involving inverse variation are nearly always easier if you think of the relation as<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=PV+=+k ALIGN=MIDDLE  ALT=\"PV+=+k\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "That is, the product of pressure and volume is constant.<br><BR>\n" );
document.write( "That makes this problem trivial.  The constant product means that if the pressure is doubled, the volume has to be cut in half.<br><BR>\n" );
document.write( "ANSWER: The volume will be cut in half.<br><BR>\n" );
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