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Heat and Thermodynamics question

2014 · Q154
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Heat and Thermodynamics question

2014 · Q154

NEETPhysicsHeat and ThermodynamicsMCQ+4 / −1
A monatomic gas at a pressure P, having a volume V expands isothermally to a volume 2V and then adiabatically to a volume 16V. The final pressure of the gas is (Take γ\gammaγ = 5/3)
  1. A
    64P
  2. B
    32P
  3. C
    P/64
  4. D
    16P
View written solutionFree

Correct answer: C

First, isothermal expansion

PV=P′(2V);P′=P2PV = P'\left( {2V} \right);P' = {P \over 2}PV=P′(2V);P′=2P​

Then, adiabatic expansion

P′(2V)γ=Pf(16V)γP'{\left( {2V} \right)^\gamma } = {P_f}{\left( {16V} \right)^\gamma }P′(2V)γ=Pf​(16V)γ
(For adiabatic process, PVγ\gamma γ = constant)

P2(2V)=Pf(16V)5/3{P \over 2}\left( {2V} \right) = {P_f}{\left( {16V} \right)^{5/3}}2P​(2V)=Pf​(16V)5/3

Pf=P2(2V16V)5/3=P2(18)5/3{P_f} = {P \over 2}{\left( {{{2V} \over {16V}}} \right)^{5/3}} = {P \over 2}{\left( {{1 \over 8}} \right)^{5/3}}Pf​=2P​(16V2V​)5/3=2P​(81​)5/3

=P2(123)5/3=P2(125)=P64 = {P \over 2}{\left( {{1 \over {{2^3}}}} \right)^{5/3}} = {P \over 2}\left( {{1 \over {{2^5}}}} \right) = {P \over {64}}=2P​(231​)5/3=2P​(251​)=64P​

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