document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1179570>Question 1179570</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The pressure P of a perfect gas varies inversely to the volume V. The pressure of a gas in a 4-L container under specific conditions is 100 atm. What is the pressure of the gas in a 45-L container under the same conditions? Use the equation P=K/V </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #809141 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13203)</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=809141\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "One very formal method: Use the given parameters to determine the value of the constant K and then use that constant with the one new parameter to determine the other new parameter.<br><BR>\n" );
document.write( "P=K/V<BR>\n" );
document.write( "100=K/4<BR>\n" );
document.write( "400=K<br><BR>\n" );
document.write( "P=K/V<BR>\n" );
document.write( "P=400/45 = 80/9<br><BR>\n" );
document.write( "For inverse variation, I think a problem is more easily solved by rewriting the formula as<br><BR>\n" );
document.write( "PV=K<br><BR>\n" );
document.write( "In that form, the relation says that the product of the pressure and the volume is a constant.  So the path to the answer to the problem is shorter:<br><BR>\n" );
document.write( "100(4)=P(45)<BR>\n" );
document.write( "P=400/45 = 80/9<br><BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );