vela said:Since T is constant, so is the righthand side of the ideal gas law. Hence you know that P1V1=P2V2. Note you don't have to find the actual volumes; you just need their ratio.
vela said:Looks good. (I'm assuming you have the right equation for the entropy. I don't know them off the top of my head.)
The entropy change for isothermal expansion of a perfect gas refers to the change in entropy that occurs when a perfect gas undergoes a process of expansion at a constant temperature.
Entropy change is important in isothermal expansion of a perfect gas because it allows us to quantify the level of disorder or randomness in the system. This can help us understand the efficiency and thermodynamic behavior of the process.
The entropy change for isothermal expansion of a perfect gas can be calculated using the equation ΔS = nRln(V₂/V₁), where ΔS is the change in entropy, n is the number of moles of gas, R is the gas constant, and V₁ and V₂ are the initial and final volumes of the gas.
The entropy change for isothermal expansion of a perfect gas can be affected by the temperature, the initial and final volumes of the gas, and the number of moles of gas present.
Yes, the entropy change for isothermal expansion of a perfect gas can be negative if the final volume of the gas is smaller than the initial volume. This means that the gas is becoming more ordered and less random, which results in a decrease in entropy.