When an ideal mono atomic gas is allowed to expand slowly until its pressure is reduced to exactly half its original values, by what factor does the volume change if the process is adiabatic.
no idea
To determine the change in volume of an ideal monoatomic gas during an adiabatic expansion process where the pressure is reduced to half its original value, we can use the relationship between pressure, volume, and temperature known as the adiabatic expansion formula.
The adiabatic expansion formula is given by:
P1 * V1^γ = P2 * V2^γ
Where:
P1 = initial pressure
V1 = initial volume
P2 = final pressure
V2 = final volume
γ = heat capacity ratio (ratio of specific heat capacities)
In the case of an ideal monoatomic gas, the value of γ is 5/3.
Let's assume the initial pressure (P1) is P and the initial volume (V1) is V.
We know that the final pressure (P2) is half the initial pressure, so P2 = P/2.
Using the adiabatic expansion formula, we can now solve for the final volume (V2) in terms of P and V:
P1 * V1^γ = P2 * V2^γ
P * V^γ = (P/2) * V2^γ
Next, we can isolate V2:
V2^γ = V^γ * P/P2
V2^γ = V^γ * P/(P/2)
V2^γ = 2 * V^γ
Taking the γth root of both sides, we get:
V2 = V * (2)^(1/γ)
For an ideal monoatomic gas, γ = 5/3, so we have:
V2 = V * (2)^(3/5)
Therefore, the volume changes by a factor of (2)^(3/5) or approximately 1.5157 during the adiabatic expansion process when the pressure is reduced to half its original value.