Thermodynamic Practice Problems And Solutions

Based on graph P-V below, what is the ratio of the work done by the gas in the process I, to the work done by the gas in the process II? Pressure (P) = 20 N/m2. Initial volume (V1) = 10 liter = 10 dm3 = 10 x 10-3 m3. Final volume (V2) = 40 liter = 40 dm3 = 40 x 10-3 m3.

Process (P) = 15 N/m2. Initial volume (V1) = 20 liter = 20 dm3 = 20 x 10-3 m3. Final volume (V2) = 60 liter = 60 dm3 = 60 x 10-3 m3. Wanted : The ratio of the work done by gas. The work done by gas in the process I :. W = P ΔV = P (V2–V1) = (20)(40-10)(10-3 m3) = (20)(30)(10-3 m3) = (600)(10-3 m3) = 0.6 m3. The work done by gas in the process II :. W = P ΔV = P (V2–V1) = (15)(60-20)(10-3 m3) = (15)(40)(10-3 m3) = (600)(10-3 m3) = 0.6 m3.

The ratio of the work done by gas in the process I and the process II :.

  • 0.6 m3 : 0.6 m3. Based on the graph below, what is the work done by helium gas in the process AB?
  • Pressure (P) = 2 x 105 N/m2 = 2 x 105 Pascal. Initial volume (V1) = 5 cm3 = 5 x 10-6 m3.

Final volume (V2) = 15 cm3 = 15 x 10-6 m3.

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Wanted : Work done by gas in process AB. W = P (V2 – V1). W = (2 x 105)(15 x 10-6 – 5 x 10-6). W = (2 x 105)(10 x 10-6) = (2 x 105)(1 x 10-5).

Based on the graph below, what is the work done in process a-b?

    • Initial pressure (P1) = 4 Pa = 4 N/m2.
    • Final pressure (P2) = 6 Pa = 6 N/m2. Initial volume (V1) = 2 m3.
    • Final volume (V2) = 4 m3. Wanted : work done I process a-b.
    • Work done by gas = area under curve a-b. W = area of triangle + area of rectangle.
    • W = ½ (6-4)(4-2) + 4(4-2). W = ½ (2)(2) + 4(2). Based on graph below, what is the work done in process A-B-C-A.
    • Work (W) = Area of the triangle A-B-C. W = ½ (20-10)(6 x 105 – 2 x 105).
    • W = ½ (10)(4 x 105). W = (5)(4 x 105). W = 2 x 106 Joule. An engine absorbs 2000 Joule of heat at a high temperature and exhausted 1200 Joule of heat at a low temperature.

What is the efficiency of the engine? Heat input (QH) = 2000 Joule. Heat output (QL) = 1200 Joule. Work done by engine (W) = 2000 – 1200 = 800 Joule.

Wanted : efficiency (e). An engine absorbs heat at 960 Kelvin and the engine discharges heat at 576 Kelvin.

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