Answer to Question #61589 in General Chemistry for Cheyenne

Question

what is the work done when a certain amount of a gas (assume the gas is ideal) expands from an initial volume of 5.0 L to a finla volume of 10.0L against a constant external pressure of 1.00atm at a constant temperature of 25 degree C? What is the internal energy change of the process? what is the heat change in this process?

Answer 1

W=-P · ΔV
1 atm = 1.01 · 105 Pa
W=1.01·10^5·((10-5)/1000)=-505 J
The minus sign means work has been done by the system.

For an ideal gas, the differential of the internal energy is given by dU = CvdT and the differential of the enthalpy is given by dH = CpdT . Since the process is isothermal, dT = 0, and so dU = 0 and dH = 0.