A cylinder containing an ideal gas has a volume of 2.6 m³ and a pressure of 1.5 × 10⁵ Pa at a temperature of 300 K.The cylinder is placed against a metal block that is maintained at 900 K and the gas expands as the pressure remains constant until the temperature of the gas reaches 900 K.The change in internal energy of the gas is +6.0 × 10⁵ J.How much heat did the gas absorb? A)0 B)3.4E+5 J C)6.0E+5 J D)1.4E+6 J E)7.8E+5 J

Question

Quizplus · Accepted Answer

The change in internal energy of the gas is given as +6.0 × 10^5 J. Since the process is isobaric (constant pressure) and the internal energy change (∆U) in an ideal gas is given by ∆U = Q - W, where Q is the heat added to the system and W is the work done by the system, we need to find W to solve for Q. The work done by the gas during an isobaric process can be calculated using W = P∆V, where P is the pressure and ∆V is the change in volume. The change in volume can be found using the ideal gas law, PV = nRT, for both initial and final states and solving for V. Since the temperature triples (from 300 K to 900 K) and the pressure remains constant, the volume must also triple (due to V ∝ T at constant P). Therefore, the volume change ∆V = 2 * 2.6 m^3 (since the volume triples, the change is twice the initial volume). The work done W = (1.5 × 10^5 Pa) * (2 * 2.6 m^3) = 7.8 × 10^5 J. Now, using ∆U = Q - W, we solve for Q: 6.0 × 10^5 J = Q - 7.8 × 10^5 J, which gives Q = 6.0 × 10^5 J + 7.8 × 10^5 J = 1.4 × 10^6 J.

A Cylinder Containing an Ideal Gas Has a Volume of 2.6