Solution Physics 213 Problem 4 Week 2 Isothermal Compression a) One and one-half moles of a diatomic gas at temperature 35oC are compressed isothermally from a volume of 0.015 m3 to a volume of 0.0015 m3. Sketch the process on a pV diagram and show what corresponds to the work done on the gas. Calculate the work done on the gas. (Ignore vibrational energy of the molecules.) The figure above shows the path along which the gas is compressed from the initial state to the final state. The path is an isotherm. The equation of state of an ideal gas is p = nRT/V. The differential of the work done by the gas is dW = pdV. Substituting for p, we obtain dW = (nRT/V) dV. Since the temperature, T, is constant (isothermal process), we can integrate the work from the initial to the final volume. Note that the work done by the gas is the negative of the work done on the gas: Wby = − Won. Vf Vf dV ⎛ V ⎞ W =−W =− ∫ pdV =−nRT ∫ = nRT ln⎜ i ⎟ on by V ⎜ V ⎟ Vi Vi ⎝ f ⎠ 3 3 Won = = (1.5 mol) (8.314 J / mol·K) (273 K + 35 K) ln(0.015 m / 0.0015 m ) = 8844 J Positive work is done on the gas. Note that we did the integral for Wby by substituting nRT/V for p and took the factor nRT out of the integral sign. This procedure is correct only for isothermal (T = constant) processes. Had the process been, for example, isobaric (constant pressure), we would have simply brought p out of the integral and found that Wby = p (Vf − Vi). The point is that the way we do the integral varies with the nature of the process. Wby = nRT ln(Vf / Vi) for isothermal processes only. b) How much heat flowed into the gas? Use the First Law and the fact that U = U(T) for ideal gases. ∆U = Qin + Won = Qin − Wby ∆U = 0 since T = constant. 0 = Qin − Wby Qin = Wby = −8844 J Note that ∆U = 0 since the temperature did not change and that for an ideal gas, the internal energy depends only on T. Heat flows out of the gas. c) What is the final pressure of the gas? p = nRT/V = (1.5 mol) × (8.314 J / mol·K) × (273 K + 35 K) / (0.0015 m3) = 2.56 × 106 Pa The initial pressure can also be found from the equation of state. p = nRT/V = (1.5 mol) × (8.314 J / mol·K) × (273 K + 35 K) / (0.015 m3) = 2.56 × 105 Pa The pressure has increased by a factor of ten while the volume has decreased by a factor of ten. The product pressure times volume is constant, since T is constant. .