Question #9802d
1 Answer
Explanation:
Your starting point here will be the fundamental thermodynamic relation, which establishes a relationship between infinitesimal changes in internal energy, entropy, and volume for a closed system at thermal equilibrium
#color(blue)( |bar(ul( color(white)(a/a)dU = TdS - PdVcolor(white)(a/a)))|)" "# , where
Now, an isothermal expansion is characterized by the fact that the temperature of the gas remains unchanged. For an ideal gas that undergoes an isothermal expansion, the fact fact that temperature is held constant implies that you have
#dU = 0#
This means that you can say
#0 = TdS - PdV implies TdS = PdV#
This can be rearranged to give
#dS = P/TdV#
Now, assuming that we're going from an initial state
#[dS = P/TdV] -> int_color(purple)(1)^color(purple)(2)#
#DeltaS = int_color(purple)(1)^color(purple)(2) P/T dV#
Since you're dealing with an ideal gas, you can use the ideal gas law equation
#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)))|)#
to say that
#P/T = (nR)/V = overbrace(nR)^(color(red)("constant")) * 1/V#
This means that you have
#DeltaS = nR * int_color(purple)(1)^color(purple)(2) 1/V dV#
Finally, this will get you
#DeltaS = nR * ln(V_2/V_1)#
Here
Plug in your values to get
#DeltaS = 0.13 color(red)(cancel(color(black)("moles"))) * "8.314 J"color(red)(cancel(color(black)("mol"^(-1))))"K"^(-1) * ln((19 color(red)(cancel(color(black)("L"))))/(11color(red)(cancel(color(black)("L")))))#
#DeltaS = color(green)(|bar(ul("0.59 J K"^(-1)))|)#
Now, does a positive change in entropy make sense?
Notice that when
That happens because the same number of molecules of gas are now floating around in a larger volume, which implies that the randomness and disorder of the system have increased.