What is the vapor pressure of CH4 at 1 atm and 298k?
Your question makes no sense to me. Vapor pressure is the pressure of a gas in equilibrium with its condensed phase (either liquid or solid) and since CH4 is a gas at 1 atm and 298 K, there is no liquid present to establish equilibrium.
To determine the vapor pressure of a substance, we often rely on empirical data or use thermodynamic calculations. In this case, we can use the Antoine equation, which relates the vapor pressure of a substance to its temperature.
The Antoine equation is given by:
log10(P) = A - (B / (T + C))
where P is the vapor pressure, T is the temperature in Kelvin, and A, B, and C are substance-specific constants.
For CH4 (methane), the Antoine constants are typically:
A = 3.337
B = 431.37
C = -40.358
Substituting these values into the equation, we can calculate the vapor pressure at a given temperature.
At 298 K:
log10(P) = 3.337 - (431.37 / (298 - 40.358))
=> P = 10^(3.337 - (431.37 / (298 - 40.358)))
Evaluating this equation will give us the vapor pressure of methane at 1 atm and 298 K.