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Which graph correctly represents the relationship between volume and temperature for an ideal gas at constant pressure and fixed mass?

Physics

ScanSolve AI Answer

Step-by-step solution

8 steps
  1. Understand the definition of angular frequency.

    Angular frequency (ω\omega) is the rate of change of phase. For a complete oscillation, the phase changes by 2π2\pi radians over a time period TT.

  2. Apply the formula for angular frequency.

    The formula for angular frequency is given by:

    ω=change in phasetime taken\omega = \frac{\text{change in phase}}{\text{time taken}}

    Given that the change in phase is 2π2\pi and the time taken is TT:

    ω=2πT\omega = \frac{2\pi}{T}
  3. Compare with the given options.

    The calculated angular frequency matches option D.

    The final answer is D) ω=2πT\boxed{\text{D) } \omega = \frac{2\pi}{T}}.


  4. Understand the question and relevant gas law.

    The question asks for the correct graph representing the relationship between volume (VV) and temperature for an ideal gas at constant pressure and fixed mass. This describes Charles's Law.

  5. State Charles's Law.

    Charles's Law states that for a fixed mass of gas at constant pressure, the volume (VV) is directly proportional to its absolute temperature (TKT_K).

    V∝TKV \propto T_K

    This can be written as V=kTKV = k T_K, where kk is a constant.

  6. Relate absolute temperature to Celsius temperature.

    Absolute temperature (TKT_K) is related to Celsius temperature (tCt_C) by the formula:

    TK=tC+273.15T_K = t_C + 273.15

    Substituting this into Charles's Law:

    V=k(tC+273.15)V = k(t_C + 273.15)

    This equation shows a linear relationship between VV and tCt_C.

  7. Analyze the graph's characteristics.

    The graph of VV versus tCt_C should be a straight line.
    When tC=−273.15∘Ct_C = -273.15^\circ C (absolute zero), the volume VV becomes zero:

    V=k(−273.15+273.15)=k(0)=0V = k(-273.15 + 273.15) = k(0) = 0

    So, the straight line graph should extrapolate to V=0V=0 at tC=−273.15∘Ct_C = -273.15^\circ C.

  8. Evaluate the given options.

    • Option A shows a straight line graph of VV versus tCt_C that extrapolates to V=0V=0 at −273∘C-273^\circ C. This is consistent with Charles's Law.
    • Option B shows a curved graph, which is incorrect for the relationship between VV and tCt_C.

    The final answer is A\boxed{\text{A}}.

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