Thermodynamic Process

Question

A 3.1-mole sample of an ideal gas is gently heated at constant temperature 290 K. The gas expands to 1.5 times its initial volume as its pressure decreases. What is the change in the internal energy of the gas? Let the ideal-gas constant R = 8.314 J/(mol • K).
 

1500 J

3000 J

-3000 J

0 J

Answer 1

An ideal gas expanding at constant temperature and quasi-statically (slowly enough to allow the system to fairly stay in thermodynamic equilibrium at all times) would be seen as a hyperbola on a Pressure-Volume diagram with equation PV = Constant.

The work done by the gas equals the change in internal energy of the gas.

Isothermal or "same-heat" expansion of the gas occurs when the gas is placed in positive thermal contact with a heat reservoir at the same temperature.

W = ∫_{(from V-initial to V-final)}PdV gives the work done by the gas; for an ideal gas and a quasi-static process, PV = nRT is applied for each point on the hyperbola.

Then W = ∫_{(from V-initial to V-final)}PdV or W = ∫_{(from V-initial to V-final)}nRTdV/V; here nRT is constant so one can write W = nRT∫_{(from V-initial to V-final)}dV/V or [nRT ln V|_{(from V-initial to V-final)}].

Placing given values, the work or internal energy of the gas is then written as (3.1 moles)(8.3145107 Joules per "Mole-Dot-Kelvin")(290 Kelvin) ln (1.5V_initial/V_initial) or 7474.745119 ln (1.5) which yields 3030.748338 Joules; this is equivalent to 3000 Joules and is positive energy or work since the gas is expanding rather than compressing.