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NCERT Exemplar · Q27

Q.One mole of a perfect gas is enclosed in a vertical cylinder of unit cross-sectional area fitted with a piston. A spring of spring constant kk and unstretched (natural) length LL connects the piston to the bottom of the cylinder, and the atmosphere above the piston exerts pressure PaP_a. Initially the spring is unstretched and the gas is in equilibrium. A certain amount of heat QQ is then supplied to the gas, and its volume increases from VoV_o to VV (the piston rises).

(a) What is the initial pressure of the gas?
(b) What is the final pressure of the gas?
(c) Using the first law of thermodynamics, write a relation between QQ, PaP_a, VV, VoV_o and kk.
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With the spring initially relaxed, force balance on the piston gives Pi=PaP_i=P_a. Raising the piston by a height equal to V−VoV-V_o (unit cross-section) stretches the spring by that amount, so the gas must additionally support the spring force k(V−Vo)k(V-V_o): Pf=Pa+k(V−Vo)P_f=P_a+k(V-V_o). The heat supplied equals the internal-energy rise plus the work done against the atmosphere and the work stored in the spring.

(a) Initial pressure

The cross-sectional area is 11. Initially the spring is unstretched, so it exerts no force. Force balance on the piston (gas pressure up, atmosphere down):

Pi⋅1=Pa⋅1 ⇒ Pi=Pa.P_i\cdot1=P_a\cdot1\ \Rightarrow\ \boxed{P_i=P_a}.

(b) Final pressure

When the volume grows from VoV_o to VV, with unit area the piston rises by a distance x=V−Vox=V-V_o. The spring, now stretched by xx, pulls the piston down with force kx=k(V−Vo)kx=k(V-V_o). Force balance now:

Pf⋅1=Pa⋅1+k(V−Vo) ⇒ Pf=Pa+k(V−Vo).P_f\cdot1=P_a\cdot1+k(V-V_o)\ \Rightarrow\ \boxed{P_f=P_a+k(V-V_o)}.

(c) First-law relation

The work done by the gas has two parts: pushing the piston against the constant atmospheric pressure, and stretching the spring: …

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