HMWK 1 - KEY.pdf - HOMEWORK 1 SOLUTIONS 1 a This is not an...

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HOMEWORK 1 SOLUTIONS 1. a. This is not an ideal gas, so start from the basic, most GENERAL equations: 𝑤 = − ∫ 𝑃𝑑𝑉 = ∫ ??? (𝑉 − ?𝑏) 𝑉2 𝑉1 𝑉2 𝑉1 𝑑𝑉 Note: n, R, T (isothermal) are constant: 𝑤 = −???[ln(𝑉 2 − ?𝑏) − ln(𝑉 1 − ?𝑏)] = − 1??? (8.314 ? (? ???) ) 200? ∗ ln 0.1 − 0.04 0.3 − 0.04 w = 2438.2 J (+ since it is compressed) = - q rev (1 st Law for an isothermal process , ΔE = q + w = 0 ) b. For an isothermal, and reversible process, Δ? = ∫ 𝑑𝑞 𝑟𝑒𝑣 𝑇 = 𝑞 𝑟𝑒𝑣 𝑇 = -w / T (from part a) = (-2438.2/200) J/(Kmol) = - 12.19 J/K ( < 0 since gas has less space to move in) c. No, the work done would be greater at constant pressure pretty easy to see from looking at the work done as the area under the PV graph: 2. a. Assuming that the values of the heat capacities remain constant over the temperature range, a. q GLASS = C T = 753 J/ (K kg) * 1kg * 100g T = 75.3 (J/K) T = q = 500J 1000 g T GLASS = 500 J/ (75.3 J/K) = 6.64 K Using the heat capacity for 500g rubber, T RUBBER = 500 J/ (188.0 J/K) = 2.66 K b. For the same transfer of heat to glass and rubber, the rubber has less than half the change in temperature
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