Question

For a fixed amount of an ideal gas P vs T plot is given as shown. Identify the incorrect option B A P(atm) straight line graph T(Kelvin) (A) The change from A to B should be isochoric (B) Volume first increases reaches maxima and then decreases (C) PV = nRT is not applicable (D) None of the statements are correct


*Provide Explanation for all the options* ​

Answer 1

Solution:

a.b,c options are incorrect.

Explanation:

please find image below.

Answer 2

Answer:

(D) None of the statements are correct.

Explanation:

For a fixed amount of an ideal gas P vs T plot is given as shown.

AB is a straight line graph.

 Pressure (atm) and Temperature (Kelvin) .

The given line does not pass through the origin.hence the volume is not constant.

     V=$(nR)(\frac{T}{P})$

 Temperature($T_{B}$) greater than the Temperature ($T_{A}$)

and the pressure t point b is( $P_{B}$) greater than pressure ($P_{A}$)

from this

   $P=bT+a$

  ⇒$\frac{P}{T} =b+\frac{a}{T}$--------(1)

Where pressure is equal to Slope of the line *temperature +constant a.

The ideal gas equation is     $PV=nRT$

                           PV directly proportional to temperature.

            i.e,  $V=\frac{T}{P}$

      and ($T_{B} > T_{A}$)

           $\frac{1}{T_{B} } < \frac{1}{T_{A} }$

  Multiplying the above equation with a then we have

         $\frac{a}{T_{B} } < \frac{a}{T_{A} }$  ------(2)

   From (1) and (2) equation we have

            $(\frac{P}{T}) _{B} < (\frac{P}{T}) _{A}$  and

            $(\frac{P}{T}) _{A} > (\frac{P}{T}) _{B}$

     Therefore    $(\frac{T}{P}) _{A} < (\frac{T}{P}) _{B}$

  Hence $V_{A} < V_{B}$ The volume increases.

(D) None of the statements are correct.

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