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QJEEAdv16Ph6c07: JEE Advanced 2016 Physics - Class 11 JEEAdv16Ph6c07 Solution: A gas is enclosed in a cylinder with a movable frictionless…

A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure Pi = 105 Pa and volume Vi = 10-3 m3 changes to a final state at Pf = (132)×105 Pa\left( {{1 \over {32}}} \right) \times {10^5}\,Pa and Vf = 8 ×\times 10-3 m3 in an adiabatic quasi-static process, such that P3V5 = constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at Pi followed by an isochoric (isovolumetric) process at volume Vf. The amount of heat supplied to the system in the two-step process is approximately

Multiple Choice Options

  • A 112 J
  • B 294 J
  • C 588 J
  • D 813 J

Step-by-Step Verified Solution

In the first process : pipiγ=pfVfγ{p_i}p_i^\gamma = {p_f}V_f^\gamma

JEE Advanced 2016 Paper 2 Offline Physics - Heat and Thermodynamics Question 60 English Explanation

⇒pipf=(VfVi)γ⇒32=8γ\Rightarrow {{{p_i}} \over {{p_f}}} = {\left( {{{{V_f}} \over {{V_i}}}} \right)^\gamma } \Rightarrow 32 = {8^\gamma }

γ=53\gamma = {5 \over 3} ... (i)

For the two step process,

W=pi(Vf−Vi)=105(7×10−3)W = {p_i}({V_f} - {V_i}) = {10^5}(7 \times {10^{ - 3}})

W=7×102JW = 7 \times {10^2}J

ΔU=f2(pfVf−piVi)\Delta U = {f \over 2}({p_f}{V_f} - {p_i}{V_i})

=1γ−1(14×102−102)= {1 \over {\gamma - 1}}\left( {{1 \over 4} \times {{10}^2} - {{10}^2}} \right)

ΔU=−32.34×102=−98×102J\Delta U = - {3 \over 2}.{3 \over 4} \times {10^2} = - {9 \over 8} \times {10^2}J

Q−W=ΔUQ - W = \Delta U

⇒Q=7×102−98×102\Rightarrow Q = 7 \times {10^2} - {9 \over 8} \times {10^2}

=478×102J=588J= {{47} \over 8} \times {10^2}J = 588J

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