How do I calculate the new pressure of liquid water using its isothermal compressibility #kappa# and expansion coefficient #alpha#?
I'm given that the isothermal compressibility #kappa = 4.7xx10^(-5) "atm"^(-1)# and expansion coefficient #alpha = 1.7xx10^(-4) "K"^(-1)# at #17^@ "C"# . The rest of the question states:
- closed rigid container filled with liquid water at an initial temperature of
#14^@ "C"# and initial pressure of #"1 atm"# .
- The temperature is changed to
#20^@ "C"# , and the new pressure is to be found.
- Assume that
#alpha# and #kappa# are negligibly different whether it's #17^@ "C"# or #14^@ "C"# or #20^@ "C"# .
What is this new pressure? I know it is #"23 atm"# , but I don't know how to do the problem.
Here are the definitions you need to know, and the calculus you may need to know:
#alpha = 1/V ((delV)/(delT))_P => alphadT_P = 1/VdV_P#
#kappa = -1/V ((delV)/(delP))_T => kappadP_T = -1/VdV_T#
#((delV)/(delP))_T# is the partial derivative of the volume with respect to the pressure at a constant temperature.#((delV)/(delT))_P# is the partial derivative of the volume with respect to the temperature at a constant pressure.
Note that #1/VdV_P# and #-1/VdV_T# are under different conditions so they are not equivalent. #"_"""_P# is constant pressure, and #"_"""_T# is constant temperature.
Also, you cannot use the ideal gas law, because it fails on liquids.
I'm given that the isothermal compressibility
- closed rigid container filled with liquid water at an initial temperature of
#14^@ "C"# and initial pressure of#"1 atm"# . - The temperature is changed to
#20^@ "C"# , and the new pressure is to be found. - Assume that
#alpha# and#kappa# are negligibly different whether it's#17^@ "C"# or#14^@ "C"# or#20^@ "C"# .
What is this new pressure? I know it is
Here are the definitions you need to know, and the calculus you may need to know:
#alpha = 1/V ((delV)/(delT))_P => alphadT_P = 1/VdV_P#
#kappa = -1/V ((delV)/(delP))_T => kappadP_T = -1/VdV_T#
#((delV)/(delP))_T# is the partial derivative of the volume with respect to the pressure at a constant temperature.#((delV)/(delT))_P# is the partial derivative of the volume with respect to the temperature at a constant pressure.
Note that
Also, you cannot use the ideal gas law, because it fails on liquids.
1 Answer
AHA! Actually, right after I wrote this question out, I figured it out.
The logic behind this is to express the change in pressure with respect to temperature at a constant volume in terms of
- The container is rigid and full of liquid water. i.e. constant volume.
- The water changes pressure due to the change in temperature. i.e.
#(dP)/(dT) = ???#
Starting from the total derivative of the differential volume
#dV = ((delV)/(delT))_P dT + ((delV)/(delP))_T dP# ,
we can divide by the partial differential temperature at a constant volume,
#cancel(((delV)/(delT))_V)^(0) = ((delV)/(delT))_Pcancel(((delT)/(delT))_V)^(1) + ((delV)/(delP))_T stackrel("goal")overbrace(((delP)/(delT))_V)#
The term that cancels to
Thus, our big equation becomes:
#((delP)/(delT))_V = (-((delV)/(delT))_P)/((delV)/(delP))_T#
If we note that the definitions of
#((delP)/(delT))_V = (cancel(-V)alpha)/(cancel(-V)kappa) = alpha/kappa#
Now, we can do this trick where we multiply out a partial differential temperature at a constant volume to turn it into an exact differential,
#dP_V = alpha/kappa dT_V#
#int_(P_1)^(P_2) dP_V = alpha/kappa int_(T_1)^(T_2) dT_V#
#P_2 - P_1 = alpha/kappa (T_2 - T_1)#
So the final expression is:
#bb(P_2 = P_1 + alpha/kappa (T_2 - T_1))#
And if we use this to evaluate the final pressure:
#color(blue)(P_2) = "1 atm" + (1.7xx10^(-4) "K"^(-1))/(4.7xx10^(-5) "atm"^(-1))("6 K")#
#= 22.7 ~~ color(blue)("23 atm")#
