Discussion Forum



1)

An ideal monoatomic gas is confined ina horizontal cylinder by a spring-loaded piston  (as shown in the figure). Initially, the gas is at temperature T1,  pressure P1  and volume V1  and the spring is in its relaxed state.  The gas is then heated very slowly to temperature T2, pressure P2  and volume V2  . during this process the piston moves out  by a distance of x, Ignoring the friction  between the piston and the cylinder, the correct statements is/are

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A) If $V_{2}=2V_{1}$ and $T_{2}=3T_{1}$ , then the energy stored in the spring is $\frac{1}{4}p_{1}V_{1}$

B) If $V_{2}=2V_{1}$ and $T_{2}=3T_{1}$ , then the change in internal energy $3p_{1}V_{1}$

C) If $V_{2}=3V_{1}$ and $T_{2}=4T_{1}$, then the work done by the gas $\frac{7}{3}p_{1}V_{1}$

D) If $V_{2}=3V_{1}$ and $T_{2}=4T_{1}$, then the heat supplied to the gas is $\frac{17}{6}p_{1}V_{1}$

Answer:

Option A,B,C

Explanation:

 Note:   This question can be solved if the right-hand side chamber is assumed to open so that its pressure remains constant even if the piston shifts towards night.

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(a)    $pV=n RT \Rightarrow p \propto\frac{T}{V}$

 Temperature  is  made three times and volume is doubled

$\Rightarrow$       $p_{2}=\frac{3}{2}p_{1}$

     Further        $x=\frac{\triangle V}{A}=\frac{V_{2}-V_{1}}{A}$

                                  $=\frac{2V_{1}-V_{1}}{A}=\frac{V_{1}}{A}$

                     $p_{2}=\frac{3p_{1}}{2}=p_{1}+\frac{kx}{A}$

$\Rightarrow$     $KX=\frac{P_{1}A}{2}$

    Energy of spring

              $\frac{1}{2}kx^{2}=\frac{p_{1}A}{4}x=\frac{p_{1}V_{1}}{4}$

(b)       $\triangle U=nc_{v}\triangle T=n\left( \frac{3}{2}R\right) \triangle T$

                           = $\frac{3}{2}(p_{2}V_{2}-p_{1}V_{1})$

                    $\frac{3}{2}\left[ \left(\frac{3}{2}P_{1}\right) (2V_{1})-p_{1}V_{1}\right]=3p_{1}V_{1}$

  (c)                        $p_{2}=\frac{4p_{1}}{3}\Rightarrow p_{2}=\frac{4}{3}p_{1}=p_{1}+\frac{kx}{A}$

            $\Rightarrow$           $kx=\frac{p_{1}A}{3}\Rightarrow x=\frac{\triangle V}{A}=\frac{2V_{1}}{A}$

                                     $W_{gas}=(p_{0}\triangle V +W_{spring})$

                  $=(p_{1}Ax+\frac{1}{2}kx.x)$

$=+\left( p_{1}A.\frac{2V_{1}}{A}+\frac{1}{2}.\frac{p_{1}A}{3}.\frac{2V_{1}}{A}\right)$

                                  $=2p_{1}V_{1}+\frac{p_{1}V_{1}}{3}=\frac{7p_{1}V_{1}}{3}$

(d)

                 $\triangle Q=W+\triangle U$

  =      $\frac{7p_{1}V_{1}}{3}+\frac{3}{2}\left( p_{2}V_{2}-p_{1}V_{1}\right)$

           =   $\frac{7p_{1}V_{1}}{3}+\frac{3}{2}\left(\frac{4}{3}p_{1}.3V_{1}-p_{1}V_{1} \right)$

               =  $\frac{7p_{1}V_{1}}{3}+\frac{9}{2}p_{1}V_{1} =\frac{41p_{1}V_{1}}{6}$

 Note:  $\triangle U=\frac{3}{2}(p_{2}V_{2}-p_{1}V_{1})$   , has been obtained in part (b).