To solve this problem without using the general gas equation, we can make use of the combined gas law, which relates the initial and final conditions of a gas sample.
The combined gas law is given by:
(P1 * V1) / (T1 * M1) = (P2 * V2) / (T2 * M2)
where: P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, T1 and T2 are the initial and final temperatures, and M1 and M2 are the molar masses of the gas.
Let's calculate the volume at 100°C and a pressure of 76.0 cm of mercury (Hg) using the given information:
Initial conditions: Mass of gas (m) = 5.6 grams Volume (V1) = 6.9 liters (dm³) Temperature (T1) = 0°C = 273.15 Kelvin Pressure (P1) = 101,325 N/m²
Final conditions: Temperature (T2) = 100°C = 373.15 Kelvin Pressure (P2) = 76.0 cm Hg = 76.0 * (101,325 N/m² / 760 mm Hg) = 10,132.24 N/m²
The molar mass (M) of the gas is not given. Since it is an ideal gas, we can use the ideal gas equation to find the molar mass using the initial conditions:
PV = nRT
R is the ideal gas constant (8.314 J/(mol·K)).
Rearranging the equation to solve for molar mass (M):
M = (m * R * T1) / (P1 * V1)
Substituting the values:
M = (5.6 g * 8.314 J/(mol·K) * 273.15 K) / (101,325 N/m² * 6.9 dm³)
M = 0.0029011 g/mol (rounded to 4 decimal places)
Now we can use the combined gas law to find the final volume (V2):
(P1 * V1) / (T1 * M1) = (P2 * V2) / (T2 * M2)
(V2 * P2) = (V1 * P1 * T2 * M1) / (T1 * M2)
V2 = (V1 * P1 * T2 * M1) / (T1 * M2 * P2)
Substituting the values:
V2 = (6.9 dm³ * 101,325 N/m² * 373.15 K * 0.0029011 g/mol) / (273.15 K * 0.0029011 g/mol * 10,132.24 N/m²)
V2 ≈ 2.339 dm³
Therefore, the volume of the gas at 100°C and a pressure of 76.0 cm Hg is approximately 2.339 liters (dm³).