Thermodynamics Derivation

Physics Problem Solution

Thermodynamics: Detailed Calculation

Derivation of $\Delta T$

Step 1: Energy & Force Relations

From energy conservation, the work done by gravity equals the change in internal energy of the gas: $$ \Delta p \Delta V = 3nR \Delta T \implies \Delta V = \frac{3nR \Delta T}{\Delta p} \quad \dots(1) $$

From the force equilibrium at the final state, the pressure difference supports the piston: $$ P_{bot} – P_{top} = \Delta p \quad \dots(2) $$

Step 2: Ideal Gas Equations

Using the ideal gas law $PV = nRT_f$ for the final state (where $T_f = T + \Delta T$): $$ P_{top} = \frac{n R T_f}{V_0 + \Delta V}, \quad P_{bot} = \frac{n R T_f}{V_0 – \Delta V} $$

Substitute these into equation (2): $$ \frac{n R T_f}{V_0 – \Delta V} – \frac{n R T_f}{V_0 + \Delta V} = \Delta p $$

Simplify the left-hand side: $$ n R T_f \left[ \frac{(V_0 + \Delta V) – (V_0 – \Delta V)}{V_0^2 – \Delta V^2} \right] = \Delta p $$ $$ n R T_f \left[ \frac{2 \Delta V}{V_0^2 – \Delta V^2} \right] = \Delta p $$ $$ 2 n R (T + \Delta T) \Delta V = \Delta p (V_0^2 – \Delta V^2) \quad \dots(3) $$

Step 3: Solving for $\Delta T$

Substitute $\Delta V$ from eq (1) into eq (3):

$$ 2 n R (T + \Delta T) \left(\frac{3 n R \Delta T}{\Delta p}\right) = \Delta p \left[ V_0^2 – \left(\frac{3 n R \Delta T}{\Delta p}\right)^2 \right] $$

Multiply both sides by $\Delta p$:

$$ 6 (nR)^2 (T + \Delta T) \Delta T = \Delta p^2 V_0^2 – (3 n R \Delta T)^2 $$

Expand terms:

$$ 6 (nR)^2 T \Delta T + 6 (nR)^2 \Delta T^2 = \Delta p^2 V_0^2 – 9 (nR)^2 \Delta T^2 $$

Rearrange to form a quadratic equation in $\Delta T$:

$$ 15 (nR)^2 \Delta T^2 + 6 (nR)^2 T \Delta T – \Delta p^2 V_0^2 = 0 $$

Divide by $(nR)^2$:

$$ 15 \Delta T^2 + 6 T \Delta T – \Delta p^2 \left(\frac{V_0}{nR}\right)^2 = 0 $$

Recall from the initial state that $p V_0 = nRT \implies \frac{V_0}{nR} = \frac{T}{p}$. Substituting this back:

$$ 15 \Delta T^2 + 6 T \Delta T – T^2 \left(\frac{\Delta p}{p}\right)^2 = 0 $$

Step 4: Final Quadratic Solution

Solve using the quadratic formula $\Delta T = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$:

$$ \Delta T = \frac{-6T \pm \sqrt{(6T)^2 – 4(15)\left(-T^2 (\frac{\Delta p}{p})^2\right)}}{2(15)} $$ $$ \Delta T = \frac{-6T \pm \sqrt{36T^2 + 60T^2 (\frac{\Delta p}{p})^2}}{30} $$

Factor out $36T^2$ from the square root (noting that $60 = 36 \times \frac{5}{3}$):

$$ \Delta T = \frac{-6T + 6T \sqrt{1 + \frac{5}{3}(\frac{\Delta p}{p})^2}}{30} $$ $$ \Delta T = \frac{6T}{30} \left[ \sqrt{1 + \frac{5}{3}\left(\frac{\Delta p}{p}\right)^2} – 1 \right] $$

Final Answer $$ \Delta T = \frac{T}{5} \left[ \sqrt{1 + \frac{5}{3}\left(\frac{\Delta p}{p}\right)^2} – 1 \right] $$

THERMAL CYU 24

Physics Problem Solution

Derivation of \(\Delta T\)