Some Special Cases of the First Law of Thermodynamics — Question 7

Question 7

The quasistatic adiabatic law follows from the first law

Use $\Delta U=Q-W$, with work by the gas positive. A supplied gas model has $pV=BT$ with $B=10.0\,\mathrm{J/K}$ and $dU=C_VdT$ with $C_V=20.0\,\mathrm{J/K}$. It expands quasistatically and adiabatically from $T_i=400$ K, $V_i=.0100\,\mathrm{m^3}$ to $V_f=.0400\,\mathrm{m^3}$. Only pressure-volume work occurs.

1. Derive the relation between $T$ and $V$ and calculate final temperature.

2. Find final pressure and work output. Derive the corresponding power law in $p$ and $V$.

Question 7 - Solution

The quasistatic adiabatic law follows from the first law

(a) Method and calculation

$$C_VdT=-p\,dV=-\frac{BT}{V}dV\ \Rightarrow\ \frac{dT}{T}=-\frac B{C_V}\frac{dV}{V}.$$$$TV^{B/C_V}=\text{constant},\quad T_f=400\left(\frac{.0100}{.0400}\right)^{1/2}=200\,\mathrm K.$$

(b) Interpretation and further calculation

$$p_f=\frac{10(200)}{.0400}=50000\,\mathrm{Pa},\quad W=-\Delta U=20(400-200)=4000\,\mathrm J.$$$$pV^{1+B/C_V}=\text{constant}\ \Rightarrow\ pV^{3/2}=\text{constant}.$$Both adiabatic insulation and the quasistatic work expression are needed for this derivation.

Physical check

The temperature decreases because positive work is funded by internal energy, with no heat input.