Osmosis and vapour pressure


Water moves into and between cells through permeable membranes. This process is driven by differences in solute concentration (sugars, acids, etc.) inside and outside the cell. This process— osmosis— creates turgor pressure and is what keeps vegetables firm and crisp. However, too much water can lead to splitting. 

Vegetables stay crisp and firm because water inside the cells is under pressure and pushes out against the cell walls and membranes. Water is drawn into the plant cells by osmosis, which ensures they stay turgid.

Osmosis is the process by which water will move through a permeable membrane (e.g. the plasmalemma inside the cell wall) from areas of low concentration of solutes to areas of higher concentration. Solutes can include dissolved sugars, acids, minerals and salts. The resulting increase in the volume of water inside the cell creates turgor pressure.

Osmosis is the process by which water moves through a permeable membrane from areas of low concentration of solutes (e.g. sugars and acids) into areas of higher concentration. This creates turgor pressure. 

When a product with a semi-permeable skin and high concentrations of solutes is placed in water, osmosis will result in it absorbing water. This can result in weight gain and improved turgidity. However, if the process goes too far, the result is swelling and cracking, as occurs with broccoli stems left in melt-water.

Vapour pressure 

Vapour pressure is the amount of water vapour held in the air. Differences in vapour pressure due to RH and temperature are termed the vapour pressure deficit (VPD). 

At any temperature and humidity, water molecules constantly change between the liquid phase—water, and the gas phase—vapour. At 50% RH water will tend to move from the liquid phase into vapour, while at 100% RH a product with high osmotic potential (eg a carrot) can absorb moisture from the air, effectively changing vapour back into liquid water. At 95−100% RH, water vapour outside most products will be in equilibrium with the water vapour inside the product, which in turn is in equilibrium with the liquid water contained in the cells. Under these conditions changes between liquid and vapour will be the same in both directions.

Water constantly moves between the liquid and vapour phase. At 100% RH the liquid and gas phases exist in a dynamic equilibrium (left). However, at lower RH, water will move out of the liquid phase and into the gas phase.


As vegetables are mostly water, the air spaces inside them are normally saturated with water vapour. While this means the internal air spaces are effectively at 100% RH, the actual amount of water held as vapour will be affected by temperature.

The amount of water vapour in the air can be expressed as vapour pressure, with units in kilopascals (KPa).

The difference in vapour pressure between the air inside and outside the product is the vapour pressure deficit (VPD). 


Calculating the vapour pressure deficit 

The VPD in KPa can be calculated from information on temperature and humidity. Large VPD values drive moisture loss, and are likely to occur when warm produce is placed in a cool room. Cooling vegetables quickly is essential to avoiding moisture loss. 

One of the easiest ways to determine vapour pressure in KPa at different combinations of temperature and humidity is by using a psychrometric chart, as shown below. This shows how quickly the amount of water vapour air can hold increases as temperature rises.

While vegetables are warm, they are almost certainly losing moisture. This is because their internal vapour pressure will be much higher than the vapour pressure inside the cool room, even if both are close to 100% RH. The difference between the vapour pressure inside the product and that in the surrounding air is termed the vapour pressure deficit (VPD).

The example shown below shows the VPD for broccoli freshly harvested at 30°C (internal RH = 100%) placed into a cool room running at 5°C and 80% RH. The vapour pressure inside the broccoli is approximately 5.2KPa, whereas that in the cool room is only 0.8KPa. The resulting vapour pressure deficit is around 4.4KPa. This difference in water vapour pressure between the inside and outside of the product has the potential to drive significant moisture loss as water moves from the broccoli into the cool room air.

If broccoli harvested at 10°C is placed into the same cool room, the vapour pressure deficit would only be around 0.7KPa (1.5KPa – 0.8KPa). Under these conditions moisture loss will be more than six times slower than in the first example.

The speed of cooling is therefore critical to reduce loss of moisture from products, even if the cool room is running at close to saturated humidity. 

Using the psychrometric chart to calculate vapour pressure deficit. An example is shown for freshly harvested broccoli at 30°C which has been placed into a cool room running at 5°C and 80% RH. The vapour pressure deficit is approximately 4.4KPa.