A spherical balloon of radius $$3\,\text{cm}$$ containing helium gas has a pressure of $$48 \times 10^{-3}\,\text{bar}$$. At the same temperature, the pressure, of a spherical balloon of radius $$12\,\text{cm}$$ containing the same amount of gas will be............ $$\times 10^{-6}\,\text{bar}$$.

We have an initial spherical balloon of radius $$r_1 = 3\;\text{cm}$$ which contains helium gas at a pressure $$P_1 = 48 \times 10^{-3}\;\text{bar}$$. A second balloon of radius $$r_2 = 12\;\text{cm}$$ is filled with the same amount of helium gas and is kept at the same temperature. For an ideal gas the relation $$PV = nRT$$ holds. Because the temperature $$T$$ and the amount of gas $$n$$ remain unchanged, the product $$PV$$ is constant, so

$$P_1 V_1 = P_2 V_2.$$

The volume of a sphere is given by the formula $$V = \dfrac{4}{3}\pi r^{3}$$. Substituting the two radii, we write

$$V_1 = \dfrac{4}{3}\pi r_1^{3}, \qquad V_2 = \dfrac{4}{3}\pi r_2^{3}.$$

Dividing the second volume by the first, the common factor $$\dfrac{4}{3}\pi$$ cancels:

$$\frac{V_2}{V_1} \;=\; \frac{r_2^{3}}{r_1^{3}}.$$

Now we substitute the numerical radii:

$$\frac{V_2}{V_1} = \left(\frac{12}{3}\right)^{3} = 4^{3} = 64.$$

Hence $$V_2 = 64\,V_1.$$ Putting this result in $$P_1 V_1 = P_2 V_2$$, we have

$$P_1 V_1 = P_2 (64\,V_1).$$

The volume factor $$V_1$$ now cancels from both sides, yielding

$$P_2 = \frac{P_1}{64}.$$

Substituting $$P_1 = 48 \times 10^{-3}\;\text{bar}$$, we get

$$P_2 = \frac{48 \times 10^{-3}}{64}\;\text{bar}.$$

We first carry out the division of the numerical coefficients. Writing the fraction explicitly,

$$\frac{48}{64} = \frac{48 \div 16}{64 \div 16} = \frac{3}{4} = 0.75.$$

Therefore,

$$P_2 = 0.75 \times 10^{-3}\;\text{bar}.$$

The answer is required in the form “$$\dots \times 10^{-6}\;\text{bar}$$”. To convert $$0.75 \times 10^{-3}$$ into a $$\times 10^{-6}$$ expression, we shift the factor $$10^{-3}$$ three powers lower:

$$0.75 \times 10^{-3} \;=\; 0.75 \times 10^{3} \times 10^{-6}.$$\; (since $$10^{-3} = 10^{3} \times 10^{-6}$$)

Now, $$0.75 \times 10^{3} = 0.75 \times 1000 = 750.$$ Hence

$$P_2 = 750 \times 10^{-6}\;\text{bar}.$$

So, the answer is $$750$$.