A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0°C. As it does so, it absorbs 208 J of heat. The values of q and w for the process will be: (R = 8.314 J / mol K) (ln 7.5 = 2.01)
q = 208 J
U = 0 at constant temperature
U = q + w
w = – q = –208 J
This problem involves a reversible isothermal expansion of an ideal gas. Let's break down the concepts step by step.
The gas expands reversibly at a constant temperature (37.0°C). This is an isothermal process. For an ideal gas undergoing an isothermal process, the internal energy change (ΔU) is zero because internal energy depends only on temperature for an ideal gas.
Mathematically:
The first law of thermodynamics states: , where:
Since , we have:
This means the heat absorbed by the system is equal in magnitude but opposite in sign to the work done on the system.
In thermodynamics, the sign convention is:
The problem states the gas absorbs 208 J of heat, so .
For a reversible isothermal expansion of an ideal gas, the work done by the gas is given by:
Where:
Plugging in the values:
First, calculate :
Then, (rounded to match the given heat value).
This confirms the work done by the system is .
From the calculation and the first law:
This matches the option: q = +208 J, w = –208 J.
First Law of Thermodynamics:
Work in Reversible Isothermal Expansion:
Internal Energy for Ideal Gas: Depends only on temperature. for isothermal processes.
One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is ws and that along the dotted line path is wd, then the integer closet to the ratio wd / ws is
