In this problem, we are tasked with calculating the pressure of an ideal gas contained in a volume of 1.6 liters, with a mass of 200 grams and a molecular mass of 28 grams per mole. The root mean square (RMS) speed of the gas molecules is given as 600 meters per second. To find the pressure, we will utilize the ideal gas law, represented by the equation:

PV = nRT

Here, P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin. To isolate P, we rearrange the equation:

P = \frac{nRT}{V}

First, we need to determine the number of moles (n). Using the relationship between mass and molar mass, we can calculate n as follows:

n = \frac{m}{M}

Substituting the values, we have:

n = \frac{200 \text{ grams}}{28 \text{ grams per mole}} = 7.14 \text{ moles}

Next, we need to find the temperature (T). The RMS speed is related to temperature and molar mass through the equation:

v_{RMS} = \sqrt{\frac{3RT}{M}}

Squaring both sides gives:

v_{RMS}^2 = \frac{3RT}{M}

Rearranging for T, we have:

T = \frac{v_{RMS}^2 \cdot M}{3R}

Substituting the known values (with molar mass converted to kilograms):

T = \frac{(600 \text{ m/s})^2 \cdot 0.028 \text{ kg/mol}}{3 \cdot 8.314 \text{ J/(mol·K)}}

Calculating this yields:

T \approx 404.1 \text{ K}

Now, we convert the volume from liters to cubic meters for consistency in units:

V = 1.6 \text{ liters} = 0.0016 \text{ m}^3

With all variables determined, we can now calculate the pressure:

P = \frac{(7.14 \text{ moles}) \cdot (8.314 \text{ J/(mol·K)}) \cdot (404.1 \text{ K})}{0.0016 \text{ m}^3}

Upon performing the calculation, we find:

P \approx 1.5 \times 10^7 \text{ Pascals}

This result represents the pressure of the ideal gas in the container. Understanding the relationships between mass, moles, temperature, and pressure is crucial in solving problems involving ideal gases.