What is the deepest place on Mars? Do humans need pressurized suits there?

Question

Is there any particular deep area, trench, fissure, lava tube or otherwise on of Mars in which a person could survive with only an oxygen supply? How deep would plants or animals have to be in Mars not to be pressurized or insulated?

Answer 1

Is there any particular deep areas of Mars in which a person could survive with only an oxygen supply without a pressurized suit?

No.

Hellas Planitia is the lowest point on Mars, the basin floor is about 7,152 m (23,465 ft) deep and the pressure is 1.16 kPa (0.168 psi). The average surface pressure of Mars is 0.6 kPa (0.087 psi). The highest point, Olympus Mons, has a height of nearly 25 km (13.6 mi or 72,000 ft) and a pressure of 0.03 kPa (0.0044 psi) - so you'd be digging over 10 miles - the deepest mine on Earth is 2.5 miles.

The Armstrong limit is 6.25 kPa (0.906 psi).

See Wikipedia's "Armstrong Limit":

"The Armstrong limit or Armstrong's line is a measure of altitude above which atmospheric pressure is sufficiently low that water boils at the normal temperature of the human body. Humans absolutely cannot survive above this limit in an unpressurized environment; above Earth, this begins 18-19 km (59,000-62,000 ft) above sea level. It is named after United States Air Force General Harry George Armstrong, who was the first to recognize this phenomenon.

At or above the Armstrong limit, exposed body fluids such as saliva, tears, urine, blood and the liquids wetting the alveoli within the lungs—but not vascular blood (blood within the circulatory system)—will boil away without a full-body pressure suit, and no amount of breathable oxygen delivered by any means will sustain life for more than a few minutes.".

Answer 2

To

Or how deep would one have to be in Mars not to need a pressurized suit?

and starting with @Rob's values and Planetery-Science.org's scale height of about 10.8 km to at least roughly ballpark an answer:

altitude (km)   pressure (kPa)
  -7.15             1.16
   0.               0.6
  25.               0.03

$$P(h) = P_0 \exp\left( -\frac{h-h_0}{h_{scale}} \right)$$

I get an altitude of -25 kilometers for the pressure to reach roughly the Armstrong Limit described in @Rob's excellent answer.

That doesn't mean that I would advocate doing so though!

edit: Based on @Uwe's comment I've extended the plot to -38 km altitude where the pressure reaches about 20 kPa, a slightly more people-friendly pressure than the absolute Armstrong limit.

note: I chose the two higher pressure points for the extrapolation, deviations from simple scale height behavior may be worse up there. Ideally one would estimate a temperature profile and use it to generate a temperature-dependent scale height as at least a step in the right direction. None the less, the answer will remain several tens of kilometers below the surface.

import numpy as np
import matplotlib.pyplot as plt

hscale    = 10.8  # km
kms, kPas = np.array([-7.15, 0.0, 25.  ]), np.array([ 1.16, 0.6,  0.03])
P0, h0    = kPas[0], kms[0]
alts      = np.arange(-38, 25.)
pressures = P0 * np.exp(-(alts - h0) / hscale)

if True:
    plt.figure()
    for i in range(2):
        plt.subplot(1, 2, i+1)
        plt.plot(kms, kPas, 'ok')
        plt.plot(alts, pressures)
        if i == 0:
            plt.yscale('log')
            plt.ylabel('pressure (kPa)', fontsize=16)
        else:
            plt.title('fitted region (linear)', fontsize=16)
            plt.xlim(-10, None)
            plt.ylim(None, 1.5)
        plt.xlabel('altitude (km)',  fontsize=16)
    plt.show()