Problem Statement
A gas undergoes a process where $p_1=1\ \text{atm}$, $V_1=10\ \text{L}$ and $p_2=4\ \text{atm}$, $V_2=4\ \text{L}$. Find the polytropic index $n$.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: For a polytropic process $pV^n = \text{const}$:
Step 2 — Apply the relevant physical law or equation: $$p_1 V_1^n = p_2 V_2^n \implies \frac{p_2}{p_1} = \left(\frac{V_1}{V_2}\right)^n$$
$$4 = \left(\frac{10}{4}\right)^n = (2.5)^n$$
Step 3 — Solve algebraically for the unknown: Taking logarithms:
Step 4 — Substitute numerical values with units: $$n = \frac{\ln4}{\ln2.5} = \frac{1.386}{0.916} = 1.513 \approx 1.5$$
Step 5 — Compute and check the result: Result: $n \approx 1.5$. (Between isothermal $n=1$ and adiabatic $n=1.4$ for diatomic gas; the gas exchanges some heat.)
Worked Calculation
$$p_1 V_1^n = p_2 V_2^n \implies \frac{p_2}{p_1} = \left(\frac{V_1}{V_2}\right)^n$$
$$4 = \left(\frac{10}{4}\right)^n = (2.5)^n$$
$$n = \frac{\ln4}{\ln2.5} = \frac{1.386}{0.916} = 1.513 \approx 1.5$$
For a polytropic process $pV^n = \text{const}$:
$$p_1 V_1^n = p_2 V_2^n \implies \frac{p_2}{p_1} = \left(\frac{V_1}{V_2}\right)^n$$
$$4 = \left(\frac{10}{4}\right)^n = (2.5)^n$$
Taking logarithms:
$$n = \frac{\ln4}{\ln2.5} = \frac{1.386}{0.916} = 1.513 \approx 1.5$$
Result: $n \approx 1.5$. (Between isothermal $n=1$ and adiabatic $n=1.4$ for diatomic gas; the gas exchanges some heat.)
Answer
$$\boxed{n = \frac{\ln4}{\ln2.5} = \frac{1.386}{0.916} = 1.513 \approx 1.5}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
Leave a Reply