Basement humidor

BuffaloTraceBuckeye

im running a cigar oasis in a 300 ct humidor I have about 250 sticks in there .  My basement stays cooler 63-64 degrees .  I have been keeping my sticks at 69 percent.  They seem a little dry and minor draw issues can I bump up the cigar oasis to about 72 percent ??

Patrickbrick

Reseason the humidor and check the calibration on the oasis, it's probably way off.  Wood is more than likely sucking the moisture out of the cigars.

BuffaloTraceBuckeye

I seasoned it the correct way 2 times.  My CO was off by about 6.8 perecrnt and I got that all figured out .  The cigars aren’t awful they just aren’t as moist as I like.  

peter4jc

Remember, humidity is referred to as 'relative' because it changes w/ the temperature.  Air that's 70% humidity at 80* has a ton more moisture in it than air that's 70% humidity at 50*.

BuffaloTraceBuckeye

I bumped it up to 72 percent to see if that helps some.   Is there a chart or anything that has the rh value at certain temperatures ? 

Echambers

Cigar-20070534 said:

I bumped it up to 72 percent to see if that helps some.   Is there a chart or anything that has the rh value at certain temperatures ? 

I don't know of any chart but you can calculate it yourself using the Clausius-Clapeyron equation. 

BuffaloTraceBuckeye

ok what is that equation????

IndustMech

The vaporization curves of most liquids have similar shapes. The vapor pressure steadily increase as the temperature increases. A good approach is to find a mathematical model for the pressure increase as a function of temperature. Experiments showed that the vapor pressure PP and temperature TT are related,

P∝e−ΔHvap/RT(1.1)(1.1)P∝e−ΔHvap/RT

where

• ΔHvapΔHvap is the Enthalpy (heat) of Vaporization and
• RR is the gas constant (8.3145 J mol-1 K-1).

If P1P1 and P2P2 are the vapor pressures at two temperatures T1T1 and T2T2, then a simple relationship can be found:

ln(P1P2)=ΔHvapR(1T2−1T1)(1.2)(1.2)ln⁡(P1P2)=ΔHvapR(1T2−1T1)

This is known as the Clausius-Clapeyron Equation and allows us to estimate the vapor pressure at another temperature, if the vapor pressure is known at some temperature, and if the enthalpy of vaporization is known.

ALTERNATIVE FORMULATION

Note the order of the temperatures in Equation 1.21.2 matters as the Clausius-Clapeyron Equation is often written with a negative sign (and switched order of temperatures):

ln(P1P2)=−ΔHvapR(1T1−1T2)ln⁡(P1P2)=−ΔHvapR(1T1−1T2)

EXAMPLE 1.11.1: VAPOR PRESSURE OF WATER

The vapor pressure of water is 1.0 atm at 373 K, and the enthalpy of vaporization is 40.7 kJ mol-1. Estimate the vapor pressure at temperature 363 and 383 K respectively.

SOLUTION

Using the Clausius-Clapeyron equation (Equation 1.21.2), we have:

P363=1.0exp[−(40,7008.3145)(1363K−1373K)]=0.697atmP363=1.0exp⁡[−(40,7008.3145)(1363K−1373K)]=0.697atm

P383=1.0exp[−(40,7008.3145)(1383K−1373K)]=1.409atmP383=1.0exp⁡[−(40,7008.3145)(1383K−1373K)]=1.409atm

Note that the increase in vapor pressure from 363 K to 373 K is 0.303 atm, but the increase from 373 to 383 K is 0.409 atm. The increase in vapor pressure is not a linear process.

Discussion

We can use the Clausius-Clapeyron equation to construct the entire vaporization curve. There is a deviation from experimental value, that is because the enthalpy of vaporization various slightly with temperature.

The Clausius-Clapeyron equation can be also applied to sublimation; the following example shows its application in estimating the heat of sublimation.

EXAMPLE 1.21.2: HEAT OF SUBLIMATION OF ICE

The vapor pressures of ice at 268 K and 273 K are 2.965 and 4.560 torr respectively. Estimate the heat of sublimation of ice.

SOLUTION

The enthalpy of sublimation is ΔHsubΔHsub. Use a piece of paper and derive the Clausius-Clapeyron equation so that you can get the form:

ΔHsub=Rln(P273P268)1268K−1273K=8.3145ln(4.5602.965)1268K−1273K=52,370Jmol−1ΔHsub=Rln⁡(P273P268)1268K−1273K=8.3145ln⁡(4.5602.965)1268K−1273K=52,370Jmol−1

Note that the heat of sublimation is the sum of heat of melting and the heat of vaporization.

IndustMech

EXERCISE 1.21.2

Show that the vapor pressure of ice at 274 K is higher than that of water at the same temperature. Note the curve of vaporization is also called the curve of evaporization.

EXAMPLE 1.31.3: HEAT OF VAPORIZATION OF ETHANOL

Calculate ΔHvapΔHvap for ethanol, given vapor pressure at 40 oC = 150 torr. The normal boiling point for ethanol is 78 oC.

SOLUTION

Recognize that we have TWO sets of (P,T)(P,T) data:

• Set 1: (150 torr at 40+273K)
• Set 2: (760 torr at 78+273K)

lnP=−ΔHvapRT+cln⁡P=−ΔHvapRT+c

Substituting into the above equation twice produces:

ln150=−ΔHvap(8.314)×(313)+cln⁡150=−ΔHvap(8.314)×(313)+c

and

ln760=−ΔHvap(8.314)×(351)+cln⁡760=−ΔHvap(8.314)×(351)+c

Subtract these two equations, to produce:

ln150−ln760−1.623=−ΔHvap8.314[1313−1351]=−ΔHvap8.314[0.0032−0.0028]ln⁡150−ln⁡760=−ΔHvap8.314[1313−1351]−1.623=−ΔHvap8.314[0.0032−0.0028]

Solving for ΔHvapΔHvap :

ΔHvap=3.90×104 joule/mole=39.0 kJ/moleΔHvap=3.90×104 joule/mole=39.0 kJ/mole

ADVANCED NOTE: 

It is important to not use the Clausius-Clapeyron equation for the solid to liquid transition. That requires the use of the more general Clapeyron equation

dPdT=ΔH¯TΔV¯dPdT=ΔH¯TΔV¯

where ΔH¯ΔH¯ and ΔV¯ΔV¯ is the molar change in enthalpy (the enthalpy of fusion in this case) and volume respectively between the two phases in the transition.

Contributors

• Chung (Peter) Chieh (Chemistry, University of Waterloo)

• Albert Censullo (California Polytechnic State University)

IndustMech

Thanks Google

Puff_Dougie

Or, you could just throw in another Bovida pack. 

BuffaloTraceBuckeye

Can I do that ?  I wasn’t sure if having a CO if they would offset each other? 

Yakster

@IndustMech, can you get me a certificate if I study and understand all that?

IndustMech

Yakster said:

@IndustMech, can you get me a certificate if I study and understand all that?

****, I only googled, copied, and pasted

Amos_Umwhat

Once upon a time, I might have followed all that.  Boy that was a long time ago.  Man I'm getting old.  

Gee THANKS guys.

YankeeMan

IndustMech said:

The vaporization curves of most liquids have similar shapes. The vapor pressure steadily increase as the temperature increases. A good approach is to find a mathematical model for the pressure increase as a function of temperature. Experiments showed that the vapor pressure PP and temperature TT are related,

P∝e−ΔHvap/RT(1.1)(1.1)P∝e−ΔHvap/RT

where

• ΔHvapΔHvap is the Enthalpy (heat) of Vaporization and
• RR is the gas constant (8.3145 J mol-1 K-1).

If P1P1 and P2P2 are the vapor pressures at two temperatures T1T1 and T2T2, then a simple relationship can be found:

ln(P1P2)=ΔHvapR(1T2−1T1)(1.2)(1.2)ln⁡(P1P2)=ΔHvapR(1T2−1T1)

This is known as the Clausius-Clapeyron Equation and allows us to estimate the vapor pressure at another temperature, if the vapor pressure is known at some temperature, and if the enthalpy of vaporization is known.

ALTERNATIVE FORMULATION

Note the order of the temperatures in Equation 1.21.2 matters as the Clausius-Clapeyron Equation is often written with a negative sign (and switched order of temperatures):

ln(P1P2)=−ΔHvapR(1T1−1T2)ln⁡(P1P2)=−ΔHvapR(1T1−1T2)

EXAMPLE 1.11.1: VAPOR PRESSURE OF WATER

The vapor pressure of water is 1.0 atm at 373 K, and the enthalpy of vaporization is 40.7 kJ mol-1. Estimate the vapor pressure at temperature 363 and 383 K respectively.

SOLUTION

Using the Clausius-Clapeyron equation (Equation 1.21.2), we have:

P363=1.0exp[−(40,7008.3145)(1363K−1373K)]=0.697atmP363=1.0exp⁡[−(40,7008.3145)(1363K−1373K)]=0.697atm

P383=1.0exp[−(40,7008.3145)(1383K−1373K)]=1.409atmP383=1.0exp⁡[−(40,7008.3145)(1383K−1373K)]=1.409atm

Note that the increase in vapor pressure from 363 K to 373 K is 0.303 atm, but the increase from 373 to 383 K is 0.409 atm. The increase in vapor pressure is not a linear process.

Discussion

We can use the Clausius-Clapeyron equation to construct the entire vaporization curve. There is a deviation from experimental value, that is because the enthalpy of vaporization various slightly with temperature.

The Clausius-Clapeyron equation can be also applied to sublimation; the following example shows its application in estimating the heat of sublimation.

EXAMPLE 1.21.2: HEAT OF SUBLIMATION OF ICE

The vapor pressures of ice at 268 K and 273 K are 2.965 and 4.560 torr respectively. Estimate the heat of sublimation of ice.

SOLUTION

The enthalpy of sublimation is ΔHsubΔHsub. Use a piece of paper and derive the Clausius-Clapeyron equation so that you can get the form:

ΔHsub=Rln(P273P268)1268K−1273K=8.3145ln(4.5602.965)1268K−1273K=52,370Jmol−1ΔHsub=Rln⁡(P273P268)1268K−1273K=8.3145ln⁡(4.5602.965)1268K−1273K=52,370Jmol−1

Note that the heat of sublimation is the sum of heat of melting and the heat of vaporization.

What he said!

Guitarded

IndustMech said:

Yakster said:

@IndustMech, can you get me a certificate if I study and understand all that?

****, I only googled, copied, and pasted

I was concerned this was just a formula in your Mensa brain waiting for the right moment to be released.