1 gram mole of an ideal gas at STP is subjected to a reversible adiaba...
Given:
- 1 gram mole of an ideal gas
- STP conditions (Standard Temperature and Pressure)
- Reversible adiabatic expansion
- Volume is doubled
- γ (gamma) = 1.4
To Find:
Change in internal energy of the gas
Explanation:
1. Ideal Gas Law:
The ideal gas law is given by the equation: PV = nRT
- P: Pressure of the gas
- V: Volume of the gas
- n: Number of moles of the gas
- R: Universal gas constant
- T: Temperature of the gas
2. STP Conditions:
STP conditions refer to a temperature of 273.15 K (0 °C) and a pressure of 1 atmosphere (atm).
3. Reversible Adiabatic Expansion:
In a reversible adiabatic expansion, there is no heat exchange between the system (gas) and its surroundings. The expansion is reversible, meaning that the gas can be brought back to its initial state without any loss of energy.
4. Adiabatic Process:
The adiabatic process is described by the equation: PV^γ = constant
- P: Pressure of the gas
- V: Volume of the gas
- γ: Ratio of specific heat capacities (Cp/Cv)
5. Relationship between γ and Change in Internal Energy:
The change in internal energy (ΔU) of an ideal gas during an adiabatic process is given by the equation: ΔU = Cv * ΔT
- ΔU: Change in internal energy
- Cv: Molar specific heat capacity at constant volume
- ΔT: Change in temperature
For an ideal gas, γ = Cp/Cv
Therefore, γ = Cp/Cv = (ΔU + PΔV)/(ΔU)
Simplifying the equation, we get: γ - 1 = PΔV / ΔU
6. Calculation:
Given that γ = 1.4 and the volume is doubled, the initial and final volumes can be represented as V1 and V2 respectively. Since the gas is in STP conditions, we can assume the initial pressure to be 1 atm.
Using the adiabatic process equation, we have:
P1V1^γ = P2V2^γ
Since the pressure is constant in this case, we can write:
V1^γ = V2^γ
(V1/V2)^γ = 1
(V2/V1)^(γ-1) = 1
Since the volume is doubled, V2 = 2V1
(2V1/V