Final answer:
The molar volume of carbon dioxide gas at 700 °C and 6500 kPa, use the Ideal Gas Law, convert temperature to Kelvin, and pressure to Pascals. After inserting the known values, we find that the molar volume is 1.2383 L/mol.
Step-by-step explanation:
The question asks for the calculation of the molar volume of carbon dioxide gas at 700 degrees Celsius and 6500 kPa. What we know initially is that standard conditions for molar volume are 0 degrees Celsius and 101.325 kPa, and the molar volume of any ideal gas under these conditions is 22.414 L. However, since the conditions given in the problem are not standard, we cannot directly use this standard molar volume.
To solve for the molar volume of CO2 at the given conditions, we can use the Ideal Gas Law, which is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. What we want is to determine the volume (V) occupied by one mole of gas, which is effectively the molar volume (Vm) under the conditions provided.
Firstly, we will need to convert the temperature from Celsius to Kelvin:
T(K) = 700 °C + 273.15 = 973.15 K
Next, insert the known values into the Ideal Gas Law rearranged to solve for Vm:
Vm = rac{RT}{P}
Where R is 8.314 J·mol−1·K−1 and P must be in Pa (Pascals).
First convert 6500 kPa to Pa:
P(Pa) = 6500 kPa × 1000 = 6.5×106 Pa
Now, calculate Vm:
Vm = rac{(8.314 J·mol−1·K−1)(973.15 K)}{(6.5×106 Pa)} = 0.0012383 m3mol−1
Convert m3 to liters:
Vm(L) = 0.0012383 m3mol−1 × 1000 L/m3 = 1.2383 L/mol
Thus, the molar volume of carbon dioxide gas at 700 °C and 6500 kPa is 1.2383 L/mol.