When a gas is compressed, its volume decreases and its temperature increases. In this scenario, we will be calculating the change in internal energy (∆U) and enthalpy (∆H) for a reversible adiabatic compression of 0.2 moles of an ideal gas from a volume of 1dm^3 to 0.25 dm^3.



The change in internal energy of a system is given by the formula:

∆U = -W

where W is the work done on the system. In a reversible adiabatic process, no heat is transferred to or from the system, so Q=0. Therefore, the first law of thermodynamics simplifies to:

∆U = -W

In this case, the work done on the system is given by:

W = -P∆V

where P is the pressure and ∆V is the change in volume. Since the process is reversible adiabatic, the ideal gas law can be used to relate the pressure, volume, and temperature of the gas:

PV^γ = constant

where γ is the ratio of specific heats for the gas. For an ideal gas, this value is equal to 1.4. Solving for P and substituting into the work formula yields:

W = -∆U = ∫PdV = -∫(constant/V^γ)dV = constant(1/Vf^(γ-1) - 1/Vi^(γ-1))

where Vi and Vf are the initial and final volumes, respectively. Substituting in the values given in the problem yields:

W = 0.2 * 8.31 * (300 - 900) * ln(0.25/1) = 4983.4 J

Therefore, the change in internal energy is:

∆U = -W = -4983.4 J



The change in enthalpy of a system is given by the formula:

∆H = ∆U + P∆V

In this case, we have already calculated ∆U. The change in volume is:

∆V = Vf - Vi = 0.25 - 1 = -0.75 dm^3

The pressure can again be calculated from the ideal gas law:

P = nRT/V = 0.2 * 8.31 * 300 / 1 = 4986 Pa

Substituting in the values yields:

∆H = ∆U + P∆V = -4983.4 + 4986 * (-0.75) = -8721.9 J

Therefore, the change in enthalpy is:

∆H = -8721.9 J



In this scenario, we have calculated the change in internal energy (∆U) and enthalpy (∆H) for a reversible adiabatic compression of 0.2 moles of an ideal gas from a volume of 1dm^3 to 0.25 dm^3. The change in internal energy