Final answer:
Using the ideal gas law, the volume for 3.00 moles of an ideal gas at 32°C and 1110 torr is calculated to be approximately 50900 milliliters after converting the temperature to Kelvin and the pressure to atmospheres.
Step-by-step explanation:
To find the volume in milliliters of 3.00 mol of an ideal gas at 32°C and a pressure of 1110 torr, we can use the ideal gas law, which is PV = nRT. Before using the equation, we need to convert the given temperature to Kelvin and the pressure to atmospheres, because the gas constant (R) value used is typically for liters, moles, Kelvin, and atmospheres.
Firstly, convert 32°C to Kelvin:
T (K) = 32°C + 273.15 = 305.15 K
Next, convert 1110 torr to atmospheres, knowing that 1 atm = 760 torr:
P (atm) = 1110 torr / 760 torr/atm = 1.4605 atm
Now we can use the gas constant R = 0.08206 L-atm/mol-K, and rearrange the ideal gas law to solve for volume (V):
V = nRT / P = 3.00 mol × 0.08206 L-atm/mol-K × 305.15 K / 1.4605 atm
After calculating, we find that the volume V ≈ 50.90 L. To convert liters to milliliters, we know that 1 L = 1000 mL, yielding:
V (mL) = 50.90 L × 1000 mL/L ≈ 50900 mL
Thus, the volume of 3.00 moles of an ideal gas under the given conditions is approximately 50900 milliliters.