A 0.250 mole sample of an ideal gas has a volume of 0.00200 m3, a pressure of 2.00 atmospheres. What is the temperature of the sample of gas?

To find the temperature of the gas, we can use the ideal gas law equation, which is PV = nRT.

Here, P = pressure = 2.00 atmospheres V = volume = 0.00200 m^3 n = number of moles = 0.250 moles R = ideal gas constant. Since the pressure is given in atmospheres and volume in m^3, we need to use the value of R = 0.0821 L.atm/(mol.K) T = temperature, which we need to find.

But before we can use these values, we need to convert the volume from m^3 to L because the gas constant R we are using is in L.atm/(mol.K).

1 m^3 = 1000 L So, V = 0.00200 m^3 = 0.00200 * 1000 L = 2 L

Now we can substitute the values into the ideal gas law equation:

2.00 atm * 2 L = 0.250 moles * R * T 4.00 L.atm = 0.250 moles * 0.0821 L.atm/(mol.K) * T 4.00 L.atm = 0.020525 L.atm/K * T

Solving for T, we get:

T = 4.00 L.atm / 0.020525 L.atm/K = 195 K

So, the temperature of the gas is 195 Kelvin.