http://www.electricalfun.com/lightning.htm

See, "The Density, Pressure, and Particle Distribution in a Lightning

Stroke near Peak Temperature" by Martin A. Uman, Richard E Orville, and Leon E Salanave in the Journal of Atmospheric Sciences, Volume 21, May, 1964, pp 306-311.

Uman, et al, estimate the average temperature  to be 24000K and a peak pressure of 18 atmospheres. 

Using a simplified basic theory involving ponderomotive forces ultimately gives

P-Pa = (10 to the -12 power) [squared current x (1 - squared (r/R))]/ (Greek P x squared R)           =>          calculated in Atmospheres provided that 

P overpressure and Pa = 1 both expressed in Atmospheres;

current in Ampéres

r: running radius from zero (on the return-stroke axis) to R: return-stroke radius in meters;

The formula gives an on-axis (max value) informatory value of P = 31 Atm (approx.) when the current is set to 1 x 10 kA and R to 1 millimeter (coaxial corona and halo excluded, of course!).

Temperature and spectrum of the plasma of your artificial lighting  can be easily and accurately measured by the alexandrite effect spectropyrometer, please refer http://www.hindawi.com/journals/aot/2010/656421/.

The pressure may be estimated by assuming that the volume of gas of the artificial lighting has not been changed at the moment when the artificial lighting occurs. The ideal gas law is pV=nkT. If the volume of the gas V is a constant, the pressure P of the artificial lighting may be written as:

                                                              P=TP' / T'

where T is the measured temperature of the artificial lighting, P' is the pressure of your experiment environment (atmosphere pressure) , and T' is the temperature of your experiment environment (room temperature). 

Article Thermal Temperature Measurements of Plasma Torch by Alexandr...

Thank you all for the insightful answers. I will look into it. 

@Yan: I see that the experiment measured the thermal temperature of a 100kW DC plasma torch. Does the same result apply to AC? I conducted three experiments using AC:

1) 6 kV, 200 kA

2) 9 kV, 300 kA

3) 12 kV, 400 kA

Moreover, do I apply the correlation (100kW = 11178 +- 382 K)  to the average active power or to the peak power. (Peak power yield too high values)

Thank you for assisting me further.

P.S: I calculated the average active power, peak power, energy and action integral (heating power).

The spectropyrometer can measure the temperature and relative spectral power distribution of any plasma torch, regardless of power and types of current of plasma sources.

The correction is due to the temperature fluctuation of the plasma torch measured. If the temperature of your plasma torch is stable, no correction is needed.

The accuracy of temperature measurement by the spectropyrometer for plasma is the same as that by the enthalpy prob, please refer http://www.liulabs.com/AnthalpyAlex.htm.

Thank you for your answer. Unfortunately, the experiment is already done and due to time constraints we cannot afford to repeat the experiment using the spectropyrometer to measure the energy. Is there a way to use your correlation (100kW = 11178 +- 382 K)  to give an theoretical estimate on the temperature that should have been generated during the experiment? thank you

Temperatures of the plasma torches depending power, the gases used, the nozzle designs, etc. Depending on particular applications, plasma torches can have lower power at high temperature, or high power at low temperature. We did not pursuit the correlation between the power and temperature.

For an artificial lighting at 10000 K, for example, assuming room temperature is 300K, the estimated pressure by my method is:

             P = 10000P' / 300 = 33.3 Atm

The pressure 33.3 Atm of the artificial lighting calculated by my method is very close to the estimated pressure 31 Atm by Francesco Lattarulo above.  

Thank you Yan for the ongoing support.

I understand from your explanation that is not feasible for me to estimate the temperatures that were present at the time of the artificial lightning shock during my experiment. Without values of temperatures it would be also not feasible for me to estimate the pressure by your method, correct?

I am sorry if I may be repeating this, but I just want to make sure. Is this approximation viable? 

1- Experiment Parameters: Experiment 1: 6 kV, 200 kA, Average Active Power = 14.11 MW, Energy = 98.79 MJ, Peak Power = 649.82 MW, Action Integral (Heating Power) = 158370.19:

2- Estimating the temperature using the correlation: 100kW = 11178 +- 382 K (I know you explained this shouldn't be universally viable, but does it work as an approximation?) and the average active power value, I end up with T = 1577467.3 K.

3- Estimating the pressure using your formula and standard room temperatures and pressures, I end up with P = 0.5360955 GPa

The temperature values seem to be too high but the pressure values are actually very reasonable. I did the same with the two other experiments with the values I mentioned a bit earlier (see attachment). Is this viable? 

Thank you in advance for taking the time to help me out and until then let's keep making the world a happier place.

The temperature of a plasma torch is not directly related to the power produced. Thus, your approximation is not tenable.  A low power plasma torch can have a high temperature, and a high power plasma torch can have a low temperature.

The highest measured temperatures of plasma torches are up to ~ 50,000 K, and your estimated temperatures of the plasma torch is not possible, as well as your estimated pressures are not possible.

Thank you Yan for the clarification. Is my question solvable without the need for an extra experiment and measurements?

I cannot think of one without need to know the temperature.

I briefly read the article entitled " The density, pressure and particle distribution in a lightling stroke near peak temperature" by Uman et al. It assumed that the lighting stroke is optically thin and achieves thermodynamic equilibrium at 24000 K. The pressure is 0.1 Atm, which is equal to the pressure at the altitude of 16000 m, and the correspond temperature at the altitude of 16000 m is about 200 K. The lighting pressure approximately estimated by my method is about:

240000 x 0.1 / 200 = 24 Atm 

This estimated lighting pressure of 24 Atm is close to 18 Atm given by Uman et al.