Thermodynamics Solution – Q25

Solution to Question 25

Problem: 1 mole of mono-atomic gas is confined between two pistons connected by a spring. The initial temperature is $T_1 = 300 \, \text{K}$. Heat is supplied until the temperature rises to $T_2 = 400 \, \text{K}$ and the spring length becomes $\eta = 1.1$ times the initial length. Find the total Heat supplied ($Q$).

1. Analysis of the System

The system consists of a gas expanding linearly against a spring. From the First Law of Thermodynamics, the total heat supplied is: $$ Q = \Delta U + W $$

Change in Internal Energy ($\Delta U$):
For a mono-atomic gas ($C_v = \frac{3}{2}R$): $$ \Delta U = n C_v (T_2 – T_1) = 1 \cdot \frac{3}{2} R (400 – 300) = 150R $$

2. Derivation of Work Done

Step A: Force & Pressure Equilibrium

Let $A$ be the area of the pistons and $x$ be the separation (spring length). The force balance on the piston (assuming equilibrium) is: $$ P_{gas} A = P_{ext} A + k(x – x_0) $$ Dividing by $A$, the pressure $P$ is a linear function of length $x$, and since volume $V = Ax$, pressure is linearly proportional to volume: $$ P = C_1 V + C_2 $$

Step B: Work as Area under P-V Graph

For a linear P-V relationship, the work done is the area of the trapezoid under the curve: $$ W = \int_{V_1}^{V_2} P \, dV = \frac{1}{2}(P_1 + P_2)(V_2 – V_1) $$

Step C: Substitution of State Variables

Given that the final spring length is $\eta$ times the initial length, the volume follows the same ratio: $V_2 = \eta V_1$. $$ W = \frac{1}{2}(P_1 + P_2)(\eta V_1 – V_1) = \frac{\eta – 1}{2} (P_1 V_1 + P_2 V_1) $$ Using the Ideal Gas Law $PV = nRT$:

$P_1 V_1 = n R T_1$
To find $P_2 V_1$, note that $P_2 V_2 = n R T_2$, so $P_2 (\eta V_1) = n R T_2 \implies P_2 V_1 = \frac{n R T_2}{\eta}$

Step D: Final Work Formula

Substituting these into the work equation: $$ W = \frac{\eta – 1}{2} \left( n R T_1 + \frac{n R T_2}{\eta} \right) $$ $$ W = \frac{n R}{2} (\eta – 1) \left( \frac{\eta T_1 + T_2}{\eta} \right) $$ $$ W = \frac{n R}{2} \left[ \frac{\eta – 1}{\eta} (\eta T_1 + T_2) \right] $$

3. Calculation

Parameters: $n=1$, $R \approx 8.314 \, \text{J/mol K}$, $T_1 = 300$, $T_2 = 400$, $\eta = 1.1$.

Work Done ($W$): $$ W = \frac{8.314}{2} \left[ \frac{0.1}{1.1} (1.1 \times 300 + 400) \right] $$ $$ W = 4.157 \times \left[ 0.0909 \times (330 + 400) \right] $$ $$ W = 4.157 \times 0.0909 \times 730 \approx 275.8 \, \text{J} $$

Internal Energy ($\Delta U$): $$ \Delta U = 1.5 \times 8.314 \times 100 = 1247.1 \, \text{J} $$

Total Heat ($Q$): $$ Q = 1247.1 + 275.8 \approx 1523 \, \text{J} $$

(Note: Rounding $R \approx 25/3$ or standard approximations yields $Q \approx 1520 \text{ J}$)

Answer: $ Q = 1520 \text{ J} $