What is the pressure of an ideal gas occupying a volume of 2 liters if its internal energy is 300 J?

1. The pressure of an ideal gas is determined by the Clapeyron-Mendeleev formula:
pV = nRT.
2. The internal energy of a monatomic gas is determined by the formula:
(since work on the gas does not take place).
U = 3/2 nRT.
3. Then the formula can be written as follows:
U = 3/2 pV.
4. From this formula we express the pressure:
p = U * 2: 3V.
5. We convert l to m3.
(in 1 liter 0.001 m3).
2 l = 0.002 m3.
6.p = 2 * 300: 3 * 0.002 = 100000 Pa = 0.1 MPa.
Answer: 0.1 MPa ideal gas pressure.