Answer:
0.938 L.
Step-by-step explanation:
The ideal gas law states that the product of pressure and volume is equal to the product of moles, the ideal gas constant and the temperature.
Since the pressure and temperature are constant and we know the volume of water vapor is 0.510 L, we can use that information to find the moles of water vapor.
n = PV/RT
The ideal gas constant R = 8.314 Latm/molK
n = 0.510 L * 1.64 atm / (8.314 Latm/molK * 373 K) = 0.001085 mol
Now we know the moles of water vapor at 373 K and 1.64 atm.
The reaction that takes place is:
CH4(g) + 2H2O(l) --> CO2(g) + 4H2(g)
This is a balanced equation, with the number of moles of each substance on both sides of the equation being equal.
So, the number of moles of methane required to convert 0.510 L of water to water vapor is 0.001085 mol/2 = 0.0005425 mol
Now we can use the ideal gas law to find the volume of methane measured at 298 K and 1.64 atm required to convert 0.510 L of water to water vapor at 373 K
V = nRT/P
V = 0.0005425 mol * 8.314 Latm/molK * 298 K / 1.64 atm = 0.938 L
Therefore, the volume of methane measured at 298 K and 1.64 atm required to convert 0.510 L of water to water vapor at 373 K is 0.938 L.