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');--grs:50px;--gtaos:105px;--gtaosi:-105px;--gtaof:210px;--gbes:calc(100% + 105px);--gbef:calc(100% + 210px);--gbet:translateX(-105px);--gbeti:translateX(105px);--vgc:repeat(12, 1fr);--gc1:repeat(12, 1fr);--gc2:repeat(6, 1fr);--gc3:repeat(4, 1fr);--gc3h:1/4;--gc4:repeat(3, 1fr);--gcm:8.333%;--gac:1fr;--gcw50:50%;--gc2tp:0 calc(var(--grs) * 2) 0 0;--gc2sw:var(--gbes);--gc2slpl:0 0 0 105px;--gc2srpf:0 105px 0 0;--gc2sof:1;--gc2sol:2;--gc50:4/span 6;--gc48a:1/5;--gc48b:5/-1;--gc84a:1/9;--gc84b:9/-1;--gc39a:1/4;--gc39b:4/-1;--gc93a:1/10;--gc93b:10/-1;--gc615a:1/7;--gc615b:8/-1;--gc516a:1/6;--gc516b:7/-1;--tppr:0;--bgp:42px;--s_cn:0;--s_ces:4px;--s_cest:4px 4px 0 0;--s_cesb:0 0 4px 4px;--s_cs:8px;--s_cm:12px;--s_cl:16px;--s_cls:16px 0 0 16px;--s_cle:0 16px 16px 0;--s_clt:16px 16px 0 0;--s_cel:28px;--s_celt:28px 28px 0 0;--s_cf:50%;--m_ds1:50ms;--m_ds2:100ms;--m_ds3:150ms;--m_ds4:200ms;--m_dm1:250ms;--m_dm2:300ms;--m_dm3:350ms;--m_dm4:400ms;--m_dl1:450ms;--m_dl2:500ms;--m_dl3:550ms;--m_dl4:600ms;--m_de1:700ms;--m_de2:800ms;--m_de3:900ms;--m_de4:1000ms;--m_el:0,0,1,1;--m_es:0.2,0,0,1;--m_esa:0.3,0,1,1;--m_esd:0,0,0,1;--m_ee:0.2,0,0,1;--m_eea:0.3,0,0.8,0.15;--m_eed:0.05,0.7,0.1,1;--m_elg:0.4,0,0.2,1;--m_elga:0.4,0,1,1;--m_elgd:0,0,0.2,1;--cc_w:255,255,255;--cc_b:0,0,0;--c_p:1,31,48;--c_op:255,255,255;--c_pc:203,230,253;--c_opc:1,31,48;--c_s:67,81,200;--c_os:255,255,255;--c_sc:223,224,255;--c_osc:0,9,101;--c_t:83,97,98;--c_ot:255,255,255;--c_tc:215,229,230;--c_otc:17,30,31;--c_e:179,38,30;--c_oe:255,255,255;--c_ec:249,222,220;--c_oec:65,14,11;--c_co:109,123,124;--c_b:255,255,255;--c_ob:17,30,32;--c_su:249,249,249;--c_osu:17,30,32;--c_suv:215,229,230;--c_osuv:60,73,74;--c_is:38,51,53;--c_iso:239,241,242;--c_ip:175,202,224;--c_sh:0,0,0;--c_st:1,31,48;--c_cov:187,201,202;--c_scr:0,0,0;--c_suchs:225,234,236;--c_such:231,239,241;--c_suc:245,245,245;--c_sucl:249,249,249;--c_scls:255,255,255;--c_sub:249,249,249;--c_sud:209,220,222}:root 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Let’s consider the reduction in vapor pressure. We can calculate the change in energy that accompanies a change in pressure using the first law of thermodynamics. If we let the symbol U represent the energy in a system and calculate the change in U that occurs when we change the volume, at constant pressure (we assume no heat is added or removed) we can write
dU represents a small change in energy, and dV represents a small change in volume. The relationship between pressure and volume, called the ideal gas law, is
where n is the number of moles of gas, R is a constant, known as the gas constant (8.31 J/mol K) and T is the temperature of the gas in kelvins. We can differentiate the ideal gas law to get dV
Combining this with the first law we get
Now, the energy required to go from the vapor pressure of the pure water in the glass, which we call the saturation vapor pressure or p0, to the vapor pressure of the water in the sponge is
The ratio p /p0 is called the water activity (aw) when we are talking about the water in the sponge or water in foods or other solids or liquids. We call it the relative humidity when we apply it to water in the air and sometimes multiply it by 100 to express it as a percent. The ratio U/n is the energy per mole of water and is called the water potential, with the symbol Ψ. Water potential has units of Joules/mole. With this substitution, we finally arrive at the equation relating the energy of the water in the sponge and its water activity
The equation means that we can express the energy state of the water in a product either as a water potential or as a water activity. Some fields of science use water potential and others use water activity. Some also use freezing point depression or osmolality, but these are all equivalent concepts. There are advantages and disadvantages to each, but the important thing to understand is that all are measures of the energy state of the water and have a strong theoretical basis. Water activity is the measure most widely used in food science and engineering with precise water activity meters.
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