Question

One mole of NH₃​ gas at 27 degrees Celsius is expanded in reversible adiabatic conditions to make the volume 8 times. Find the temperature and work done.

(a) Temperature increases and work is done on the gas.
(b) Temperature decreases and work is done on the gas.
(c) Temperature increases and work is done by the gas.
(d) Temperature decreases and work is done by the gas.

Answer 1

The temperature after adiabatic expansion of NH₃ gas and the work done can be calculated using the equations for a reversible adiabatic process. The final temperature is derived from the relation involving the initial and final volume and temperature, and the work done is computed from the change in internal energy.

Step-by-step explanation:

The question pertains to the adiabatic expansion of ammonia gas (NH₃). In a reversible adiabatic process, there is no heat exchange with the surroundings. The work done is related to the change in internal energy of the gas and, hence, its temperature. For an adiabatic process involving a perfect gas, the following equation holds, assuming the heat capacity ratio (γ = Cp/Cv) is known:

PV^γ = constant

Additionally, we have the relation between initial and final temperatures (T1 and T2) and volumes (V1 and V2):

T1V1^γ-1 = T2V2^γ-1

Given that the initial temperature (T1) is 27°C (or 300K), V2 is 8 times V1, and ammonia has a γ value of approximately 1.3:

(300K)V1^1.3-1 = T2(V1*8)^(1.3-1)

This allows for the calculation of T2. The work done (W) is equal to the change in internal energy (ΔU), and since the process is adiabatic, ΔU = -W (assuming the work is done by the gas).

W = nCvΔT = nCv(T2 - T1)

Putting values into this equation gives the work done by the gas.