The internal energy of a monatomic ideal gas in a cylinder with a volume of 0.02 m3

The internal energy of a monatomic ideal gas in a cylinder with a volume of 0.02 m3 is equal to 600 J. Determine the gas pressure in kilopascals.

To calculate the pressure of the specified ideal gas, we use the formula: U = i * P * V / 2, whence P = 2 * U / (i * V).

Data: U – internal energy (U = 600 J); i is the number of degrees of freedom of molecules (i = 3); V is the volume of the cylinder (V = 0.02 m3).

Let’s make a calculation: P = 2 * U / (i * V) = 2 * 600 / (3 * 0.02) = 20 * 10 ^ 3 Pa.

Answer: The pressure of the indicated ideal gas in the cylinder is 20 kPa.