Question #732f1
1 Answer
No, you will not.
Explanation:
First of all, the units for the ideal gas constant,
#8.314 "J"/("mol" * "K")#
That said, you can prove that this value of
#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#
where
Rearrange to solve for
#R = (PV)/(nT)#
Notice that the units given to you for
#R = ["J"/("mol" * "K")]#
Compare this with the above equation to see that the units for pressure and volume must multiply to give you joules,
Since you know that
#"1 J" = "1 N" xx "1 m"#
you can say that the product of the pressure and the volume will be expressed in
#P * V = "J" = "N" * "m"#
Now, if you were to express the volume in cubic meters and the pressure in pascals,
#P * V = "N" * "m" = overbrace("N"/"m"^2)^(color(blue)("= Pa")) * "m"^3 = "Pa" * "m"^3#
Now, you can express the volume in liters,
#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 L" = 10^(-3)"m"^3)color(white)(a/a)|)))" "# and#" "color(purple)(|bar(ul(color(white)(a/a)color(black)("1 kPa" = 10^(3)"kPa")color(white)(a/a)|)))#
to get
#P * V = color(red)(cancel(color(black)(10^3)))"kPa" * color(red)(cancel(color(black)(10^(-3))))"L" = "kPa" * "L"#
Therefore, in order to have
#R = 8.314 "J"/("mol" * "K")#
you need to have
#R = 8.314("kPa" * "L")/("mol" * "K")#
This means that using this value of
In fact, you can use the following conversion factors
#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 atm " = " 760 mmHg")color(white)(a/a)|)))" "# and#" "color(purple)(|bar(ul(color(white)(a/a)color(black)("1 atm " = "101.325 kPa")color(white)(a/a)|)))#
to find the correct value of
You will have
#8.314 (color(red)(cancel(color(black)("kPa"))) * "L")/("mol" * "K") * (1color(red)(cancel(color(black)("atm"))))/(101.325color(red)(cancel(color(black)("kPa")))) * "760 mmHg"/(1color(red)(cancel(color(black)("atm")))) = 62.36("mmHg" * "L")/("mol" * "K")#
Therefore, the correct value of
#color(green)(|bar(ul(color(white)(a/a)color(black)(R = 62.36("mmHg" * "L")/("mol" * "K"))color(white)(a/a)|)))#