1. Frequency formula for an open organ pipe

For a pipe open at both ends, the allowed harmonics are $f_n = \frac{n v}{2L}, \quad n=1,2,3,\dots$

So:

• Fundamental frequency: $f_1 = \frac{v}{2L}$
• Second harmonic: $f_2 = \frac{2v}{2L} = \frac{v}{L}$

Hence, the frequency difference is $f_2 - f_1 = \frac{v}{L} - \frac{v}{2L} = \frac{v}{2L}$

So the difference equals the fundamental frequency.

2. Speed of sound in the given gas

For gases under similar conditions, $v \propto \frac{1}{\sqrt{\rho}}$

Given the gas has double the density of air, $\rho_{gas} = 2\rho_{air}$

Therefore, $v_{gas} = \frac{v_{air}}{\sqrt{2}} = \frac{300}{\sqrt{2}} \text{ m/s}$

$v_{gas} \approx 212.13 \text{ m/s}$

3. Use pipe length

Given: $L = 1 \text{ m}$

Thus, $f_2 - f_1 = \frac{v}{2L} = \frac{212.13}{2 \times 1}$

$f_2 - f_1 \approx 106.07 \text{ Hz}$

4. Final answer

The required frequency difference is $\boxed{106.07 \text{ Hz}}$

For an integer-type response, this is approximately $\boxed{106}$