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11. thermodynamics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics11. thermodynamics1 Mark
MCQ
One mole of ideal gas undergoes a linear process as shown in the figure below. Its temperature expressed as a function of volume $V$ is
  • A
    $\frac{p_0 V_0}{R}$
  • B
    $\frac{p_0 V}{R}$
  • ✓
    $\frac{p_0 V}{R}\left(1-\frac{V}{V_0}\right)$
  • D
    $\frac{p_0 V_0}{R}\left(1-\left(\frac{V}{V_0}\right)^2\right)$
✓

Answer

Correct option: C.
$\frac{p_0 V}{R}\left(1-\frac{V}{V_0}\right)$
c
(c)

Process given is

To find process equation, we use two point form of equation of straight line,

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

Here, $(x, y,)=(0, p_0$, and $\left(x_2, y_2\right)=\left(V_0, 0\right)$ Process equation is

$p=p_0-\frac{p_0}{V_0} \cdot V$

As, $\quad p=\frac{R T}{V}$

$\Rightarrow \frac{R T}{V}=p_0-\frac{p_0}{V_0} \cdot V$

$\Rightarrow T=\frac{p_0 V}{R}\left(1-\frac{V}{V_0}\right)$

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