Kinetic Theory of Gases: Practice Problem & Solution
In a thermodynamic process, pressure (in Pa) varies as, \(P = a + bV\) [where \(V\) represents volume in \(\text{m}^3\), \(a\) and \(b\) are constants]. The work done by gas during expansion from \(2 \text{m}^3\) to \(3 \text{m}^3\) will be
Solution Explained:
To solve this problem, we apply the core principles of Kinetic Theory of Gases. Understanding the underlying formula is key to arriving at the correct answer below:
Work done \(W = \int_{V_1}^{V_2} P dV = \int_2^3 (a+bV) dV = \left[aV + \frac{bV^2}{2}\right]_2^3 = a(3-2) + \frac{b}{2}(9-4) = a + \frac{5b}{2}\).
Leave a Reply