Question

Two moles of nitrogen gas, with (γ=7/5) for ideal diatomic gases, occupies a volume of (10⁻² m³) in an insulated cylinder at temperature (300 K). The gas is adiabatically and reversibly compressed to a volume of (5 L). The piston of the cylinder is locked in its place, and the insulation around the cylinder is removed. The heat-conducting cylinder is then placed in a (300-K) bath. Heat from the compressed gas leaves the gas, and the temperature of the gas becomes (300 K) again. The gas is then slowly expanded at the fixed temperature (300 K) until the volume of the gas becomes (10⁻² m³), thus making a complete cycle for the gas. For the entire cycle, calculate

a) the work done by the gas,
b) the heat into or out of the gas,
c) the change in the internal energy of the gas, and
d) the change in entropy of the gas.

Answer 1

To calculate the work done, heat transfer, change in internal energy, and change in entropy of the nitrogen gas during a cycle, we can use the formulas for adiabatic processes and the first law of thermodynamics.

Step-by-step explanation:

In the given scenario, a cycle is formed by compressing and expanding the nitrogen gas. Let's calculate the values for each part of the cycle:

1. (a) To find the work done by the gas, we can use the formula for reversible adiabatic compression: W = -ΔU, where W is the work, and ΔU is the change in internal energy. Since the process is adiabatic, there is no heat transfer, so the work done by the gas is equal to the change in internal energy. Therefore, W = ΔU.
2. (b) The heat into or out of the gas can be calculated using the first law of thermodynamics: Q = ΔU + W.
3. (c) The change in internal energy of the gas can be found using the equation ΔU = Q - W.
4. (d) The change in entropy of the gas can be calculated using the formula ΔS = Q / T, where ΔS is the change in entropy, Q is the heat transfer, and T is the temperature.

Now, we can substitute the given values and calculate the answers for each part of the cycle.