The temperature of a gas having $$2.0 \times 10^{25}$$ molecules per cubic meter at $$1.38$$ atm (Given, $$k = 1.38 \times 10^{-23} \text{ J K}^{-1}$$) is:

To find the temperature of a gas given its number density and pressure, we start with the ideal gas law in the form $$P = nkT$$, where $$n$$ is the number density (molecules per unit volume), $$k$$ is Boltzmann's constant, and $$T$$ is the temperature.

Converting the given pressure of 1.38 atm to SI units gives $$P = 1.38$$ atm $$= 1.38 \times 1.01325 \times 10^5$$ Pa $$= 1.398 \times 10^5$$ Pa.

Solving for temperature yields $$T = \frac{P}{nk} = \frac{1.398 \times 10^5}{2.0 \times 10^{25} \times 1.38 \times 10^{-23}}$$ and $$= \frac{1.398 \times 10^5}{2.76 \times 10^2} = \frac{1.398 \times 10^5}{276} \approx 506.5$$ K $$\approx 500$$ K.

The correct answer is Option (1): 500 K.