Physic Labs

Problem 2

One mole of a monatomic ideal gas, initially at absolute temperature T1T_1, is heated to T2>T1T_2>T_1 in a constant-pressure process. Let RgR_g be the gas constant; assume thermodynamic equilibrium throughout and no phase change. Define WW as work done by the gas and Q>0Q>0 as heat absorbed, so the first law is ΔU=Q−W\Delta U=Q-W. (a) Use the equation of state to find the volume change and work. (b) Calculate the internal-energy change from the molecular degrees of freedom. (c) Determine the heat transfer, check its sign, and explain how it is divided between raising internal energy and doing work.
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CV=32Rg(monoatomic),ΔU=nCVΔTC_V=\frac32R_g\quad(\text{monoatomic}),\quad \Delta U=nC_V\Delta T
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A monatomic gas has three translational degrees of freedom, so CV=3Rg/2C_V=3R_g/2 and internal energy changes with temperature. Pressure remains constant and the gas follows the ideal-gas model.