The water activity of a food equals the fractional relative humidity of air in equilibrium with that food. At the same specified temperature, how does the food’s water-vapor partial pressure compare with that in the equilibrium air?
At temperature $T$,water activity is $a_w=p_{\mathrm{food}}/p_{\mathrm{sat}}(T)$ and fractional relative humidity is $p_{\mathrm{air}}/p_{\mathrm{sat}}(T)$. Equality of these ratios, with the same saturation pressure, gives $p_{\mathrm{food}}=p_{\mathrm{air}}$. Equal vapor pressures also mean no net vapor transfer at equilibrium. **A**.
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