F 7. Half a Mole of an Ideal Monatomic Gas Expansion

Problem:

Half a mole of an ideal monatomic gas at a pressure of 425 kPa and a temperature of 300 K expands until the pressure has diminished to 150 kPa

Part (a) - Isothermal Expansion:

For an isothermal expansion, use ideal gas law and work done is equal to heat absorbed. For an adiabatic expansion, use adiabatic equations and heat absorbed is zero. So answer is 0.

In this problem, we are given the initial pressure (P1) of half a mole of an ideal monatomic gas (Equation of state: PV = nRT), which is 425 kPa, and the initial temperature (T1) of 300 K. The gas expands until the final pressure (P2) is 150 kPa.

  1. To find the final temperature, we can rearrange the ideal gas law equation to T = PV / (nR). Substitute the given values to find T.
  2. To find the final volume, we can rearrange the equation P1V1 = P2V2. Substitute the given values to find V2.
  3. The work done can be calculated using the equation W = PΔV, where P is the pressure difference and ΔV is the change in volume. Substitute the given values to find W.
  4. The heat absorbed can be calculated using the equation Q = W, since the expansion is isothermal and there is no change in internal energy. Therefore, Q = W.

Part (b) - Adiabatic Expansion:

For an adiabatic expansion, use the equation P1V1^γ = P2V2^γ, where γ is the adiabatic index or heat capacity ratio. We can also use the equation P1V1^(γ-1)T1 = P2V2^(γ-1)T2, where T1 and T2 are the initial and final temperatures.

What are the final temperature and volume, work done, and heat absorbed by the gas in an isothermal expansion? The final temperature and volume can be found by rearranging equations and substituting values. The work done is calculated using the pressure difference and change in volume, while the heat absorbed is equal to the work done in an isothermal expansion.
← Describe the formation of an aqueous libr solution when solid libr dissolves in water Calculating oxygen reacted in chemical reaction →

More Posts: