How do I calculate the new pressure of liquid water using its isothermal compressibility kappaκ and expansion coefficient alphaα?
I'm given that the isothermal compressibility kappa = 4.7xx10^(-5) "atm"^(-1)κ=4.7×10−5atm−1 and expansion coefficient alpha = 1.7xx10^(-4) "K"^(-1)α=1.7×10−4K−1 at 17^@ "C"17∘C . The rest of the question states:
- closed rigid container filled with liquid water at an initial temperature of
14^@ "C"14∘C and initial pressure of "1 atm"1 atm .
- The temperature is changed to
20^@ "C"20∘C , and the new pressure is to be found.
- Assume that
alphaα and kappaκ are negligibly different whether it's 17^@ "C"17∘C or 14^@ "C"14∘C or 20^@ "C"20∘C .
What is this new pressure? I know it is "23 atm"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_Pα=1V(∂V∂T)P⇒αdTP=1VdVP
kappa = -1/V ((delV)/(delP))_T => kappadP_T = -1/VdV_Tκ=−1V(∂V∂P)T⇒κdPT=−1VdVT
((delV)/(delP))_T(∂V∂P)T is the partial derivative of the volume with respect to the pressure at a constant temperature.((delV)/(delT))_P(∂V∂T)P is the partial derivative of the volume with respect to the temperature at a constant pressure.
Note that 1/VdV_P1VdVP and -1/VdV_T−1VdVT are under different conditions so they are not equivalent. "_"""_P_P is constant pressure, and "_"""_T_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"14∘C and initial pressure of"1 atm"1 atm . - The temperature is changed to
20^@ "C"20∘C , and the new pressure is to be found. - Assume that
alphaα andkappaκ are negligibly different whether it's17^@ "C"17∘C or14^@ "C"14∘C or20^@ "C"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_Pα=1V(∂V∂T)P⇒αdTP=1VdVP
kappa = -1/V ((delV)/(delP))_T => kappadP_T = -1/VdV_Tκ=−1V(∂V∂P)T⇒κdPT=−1VdVT
((delV)/(delP))_T(∂V∂P)T is the partial derivative of the volume with respect to the pressure at a constant temperature.((delV)/(delT))_P(∂V∂T)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) = ???dPdT=???
Starting from the total derivative of the differential volume
dV = ((delV)/(delT))_P dT + ((delV)/(delP))_T dPdV=(∂V∂T)PdT+(∂V∂P)TdP ,
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")
