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Name the THREE basic gas processes..

Physics

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QUESTION 4: STRESS, STRAIN AND YOUNG MODULUS (continuation)

4.3.2 The strain.

  1. Identify the given values.

    Original length (L0L_0) = 500 mm = 0.5 m
    Elongation (ΔL\Delta L) = 0.25 mm = 0.00025 m

  2. Calculate the strain (ϵ\epsilon).

    ϵ=ΔLL0\epsilon = \frac{\Delta L}{L_0} ϵ=0.00025 m0.5 m\epsilon = \frac{0.00025 \text{ m}}{0.5 \text{ m}} ϵ=0.0005\epsilon = 0.0005

    The strain is 0.0005\boxed{0.0005}.

    4.3.3 The Young modulus of the material.

  3. Identify the given and calculated values.

    Stress (σ\sigma) = 101.86 MPa = 101.86×106101.86 \times 10^6 Pa (from 4.3.1)
    Strain (ϵ\epsilon) = 0.0005 (from 4.3.2)

  4. Calculate the Young's Modulus (EE).

    E=σϵE = \frac{\sigma}{\epsilon} E=101.86×106 Pa0.0005E = \frac{101.86 \times 10^6 \text{ Pa}}{0.0005} E=203.72×109 PaE = 203.72 \times 10^9 \text{ Pa} E=203.72 GPaE = 203.72 \text{ GPa}

    The Young modulus of the material is 203.72 GPa\boxed{203.72 \text{ GPa}}.

    QUESTION 5: HEAT

    5.1 Name the THREE basic gas processes.

    • Isobaric process (constant pressure)
    • Isochoric process (constant volume)
    • Isothermal process (constant temperature)

    5.2 A square metal plate with sides 1 m long is at a temperature of 31 °C. There is a hole of 600 mm in diameter in the centre. The linear coefficient of expansion of the metal is 12.5×10−612.5 \times 10^{-6} /K.

    5.2.1 Calculate: The temperature of the metal plate in if it is heated until the sides are 1,005 m long.

  5. Identify the given values.

    Initial length (L0L_0) = 1 m
    Final length (LfL_f) = 1.005 m
    Initial temperature (T0T_0) = 31 °C
    Linear coefficient of expansion (α\alpha) = 12.5×10−612.5 \times 10^{-6} /K

  6. Calculate the change in temperature (ΔT\Delta T).

    The formula for linear expansion is ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T.

    ΔT=Lf−L0αL0\Delta T = \frac{L_f - L_0}{\alpha L_0} ΔT=1.005 m−1 m(12.5×10−6 /K)×1 m\Delta T = \frac{1.005 \text{ m} - 1 \text{ m}}{(12.5 \times 10^{-6} \text{ /K}) \times 1 \text{ m}} ΔT=0.00512.5×10−6=400 K\Delta T = \frac{0.005}{12.5 \times 10^{-6}} = 400 \text{ K}
  7. Calculate the final temperature (TfT_f).

    Tf=T0+ΔTT_f = T_0 + \Delta T Tf=31 °C+400 °C=431 °CT_f = 31 \text{ °C} + 400 \text{ °C} = 431 \text{ °C}

    The final temperature of the metal plate is 431 °C\boxed{431 \text{ °C}}.

    5.2.2 The diameter of the hole at this temperature.

  8. Identify the given and calculated values.

    Initial hole diameter (D0D_0) = 600 mm = 0.6 m
    Linear coefficient of expansion (α\alpha) = 12.5×10−612.5 \times 10^{-6} /K
    Change in temperature (ΔT\Delta T) = 400 K (from 5.2.1)

  9. Calculate the final diameter (DfD_f).

    The hole expands by the same linear expansion formula.

    Df=D0(1+αΔT)D_f = D_0 (1 + \alpha \Delta T) Df=0.6 m(1+(12.5×10−6 /K)×400 K)D_f = 0.6 \text{ m} (1 + (12.5 \times 10^{-6} \text{ /K}) \times 400 \text{ K}) Df=0.6 m(1+0.005)D_f = 0.6 \text{ m} (1 + 0.005) Df=0.6 m×1.005D_f = 0.6 \text{ m} \times 1.005 Df=0.603 mD_f = 0.603 \text{ m}

    The diameter of the hole at this temperature is 0.603 m\boxed{0.603 \text{ m}}.

    5.3 A cylinder has a capacity of 0,075 m3^3 and is filled with oxygen at a pressure of 380 kPa and a temperature of 50 °C. Later it is found that owing to a leak, the pressure has dropped to 280 kPa and the temperature has decreased to 25 °C.

    5.3.1 Calculate the mass of the oxygen that was initially in the cylinder if the gas constant is 265 J/kg.K.

  10. Identify the initial conditions and convert units.

    Volume (VV) = 0.075 m3^3
    Initial pressure (P1P_1) = 380 kPa = 380×103380 \times 10^3 Pa
    Initial temperature (T1T_1) = 50 °C = 50+273.15=323.1550 + 273.15 = 323.15 K
    Gas constant (RR) = 265 J/kg.K

  11. Use the ideal gas law to calculate the initial mass (m1m_1).

    The ideal gas law is PV=mRTPV = mRT.

    m1=P1VRT1m_1 = \frac{P_1 V}{R T_1} m1=(380×103 Pa)×(0.075 m3)(265 J/kg.K)×(323.15 K)m_1 = \frac{(380 \times 10^3 \text{ Pa}) \times (0.075 \text{ m}^3)}{(265 \text{ J/kg.K}) \times (323.15 \text{ K})} m1=2850085634.75m_1 = \frac{28500}{85634.75} m1≈0.3328 kgm_1 \approx 0.3328 \text{ kg}

    The mass of the oxygen initially in the cylinder is 0.3328 kg\boxed{0.3328 \text{ kg}}.

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