Problem 2

One mole of monatomic ideal gas initially at absolute temperature

T_i

and pressure

p

is heated slowly at constant pressure to

T_f>T_i

. It obeys

pV=R_gT

, where

R_g

is the gas constant; neglect macroscopic kinetic and potential energy changes. Take work done by the gas and heat entering it as positive. (a) Find the volume change and work done by the gas. (b) Find the change in internal energy. (c) Find the heat added and the ratio of heat to work.

problems.proof.stepOf

problems.proof.prevStep problems.proof.nextStep

\frac QW=\frac52,\quad \boxed{W=R_g\Delta T,\;\Delta U=\frac32R_g\Delta T,\;Q=\frac52R_g\Delta T}

problems.proof.analysis

Dividing heat by isobaric work cancels $R_g\Delta T$ and gives $5/2$ . Each energy has joule units because $R_g\Delta T$ is energy per mole for one mole.

problems.proof.pitfall. Keep the sign convention consistent: if work done by the gas is positive, use

Q=\Delta U+W

, not

Q=\Delta U-W