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Step-by-Step Solution
Step 1: Identify the Given Quantities
• Mean kinetic energy of one molecule, $E = 4 \times 10^{-14}\,\text{erg}$.
• Pressure of the gas, $P$, is equivalent to the pressure due to a column of mercury of height $h = 2\,\text{cm}$.
• Density of mercury, $\rho = 13.6\,\text{g/cm}^3$.
• Acceleration due to gravity, $g = 980\,\text{cm/s}^2$.
• Volume of the gas, $V = 4\,\text{cm}^3$.
• We need to find the number of molecules, $N$.
Step 2: Relate Mean Kinetic Energy to Temperature
For a monoatomic ideal gas, the mean kinetic energy of a molecule is given by:
$$E = \frac{3}{2}kT.$$
Hence, the temperature $T$ can be expressed as:
$$T = \frac{2E}{3k}.$$
Here, $k$ is the Boltzmann constant.
Step 3: Write the Ideal Gas Equation in Terms of Boltzmann’s Constant
The ideal gas equation in terms of the number of molecules $N$ is:
$$P\,V = N\,k\,T.$$
Substitute $T = \frac{2E}{3k}$ from Step 2:
$$P\,V = N\,k \,\left(\frac{2E}{3k}\right).$$
Simplifying, we get:
$$P\,V = \frac{2}{3}\,E \,N.$$
Step 4: Calculate the Gas Pressure $P$ in cgs Units
We are given that the gas pressure $P$ corresponds to a mercury column of height $h = 2\,\text{cm}$. In cgs units:
$$P = \rho\,g\,h = (13.6\,\text{g/cm}^3)\,(980\,\text{cm/s}^2)\,(2\,\text{cm}).$$
Multiplying these:
$$P = 13.6 \times 980 \times 2 \,\text{dyn/cm}^2.$$
For simplicity, we will carry the symbolic product forward and insert approximate values as needed.
Step 5: Substitute All Values into $P V = \frac{2}{3} E\,N$
1. Left-hand side (LHS):
$$P\,V = (\rho\,g\,h)\,(4\,\text{cm}^3).$$
Numerically,
$$P\,V = 13.6 \times 980 \times 2 \,\text{dyn/cm}^2 \times 4\,\text{cm}^3.$$
2. Right-hand side (RHS):
$$\frac{2}{3} E\,N = \frac{2}{3} \times (4 \times 10^{-14}\,\text{erg})\,N.$$
Thus, equating LHS and RHS:
$$\rho\,g\,h \times 4 = \frac{2}{3} E\,N.$$
Step 6: Solve for the Number of Molecules $N$
Rearrange to find $N$:
$$N = \frac{3}{2E}\,\rho\,g\,h \times 4.$$
Substitute values carefully (keeping track of powers of 10 if needed). The detailed calculation shows this product leads to a value near:
$$N \approx 4.0 \times 10^{18}.$$
Final Answer
The number of molecules in the given volume under the specified conditions is
$$\boxed{4.0 \times 10^{18}}.$
