What pressure is exerted by 0.010 mol of methane gas if it is volume is 0.26 L at 373°K?
PV=nRT Solve for pressure.
To find the pressure exerted by the methane gas, we can use the ideal gas law, which states that:
PV = nRT
Where:
- P is the pressure in atmospheres (atm)
- V is the volume in liters (L)
- n is the number of moles of the gas
- R is the ideal gas constant (0.0821 L·atm/(K·mol))
- T is the temperature in Kelvin (K)
Given:
- n = 0.010 mol
- V = 0.26 L
- T = 373 K
We can plug these values into the equation and solve for P.
P * 0.26 L = 0.010 mol * 0.0821 L·atm/(K·mol) * 373 K
Simplifying:
P * 0.26 L = 0.010 mol * 30.5441 L·atm
P * 0.26 L = 0.3054 L·atm
P = 0.3054 L·atm / 0.26 L
P ≈ 1.17 atm
Therefore, the pressure exerted by 0.010 mol of methane gas with a volume of 0.26 L at 373 K is approximately 1.17 atm.