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First Law of Thermodynamics

A gas is encl...

Question

A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure $P_{i} = 10^{5}$ Pa and volume $V_{i} = 10^{- 3} m^{3}$ changes to a final state at $P_{f} = (1 / 32) \times 10^{5}$ Pa and $V_{f} = 8 \times 10^{- 3} m^{3}$ in an adiabatic quasi-static process, such that $P^{3} V^{5} =$ 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 $P_{i}$ followed by an isochoric (isovolumetric) process at volume $V_{f}$ . The amount of heat supplied to the system in the two-step process is approximately.

112

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294

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588

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813

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Solution

The correct option is D $813$ J
for adiabatic process
$P^{3} V^{3} = C$
$P V^{5 / 3} = K$
$r = 5 / 3$
$w_{a d i a b a t i c} = \frac{1}{1 - r} (P_{2} V_{2} - P_{1} V_{1})$
$= \frac{1}{1 - \frac{5}{3}} (\frac{1}{32} \times 10^{5} \times \frac{8}{10^{3}} - 10^{5} \times \frac{1}{10^{3}})$
$= 112.5 J$
$W_{i s o b a r i c} = P Δ V$
$= 10^{5} \times (8 - 1) \times 10^{- 3}$
$= 7 \times 100$
$= 700 J$
$\Rightarrow W_{a d i a b a t i c} + W_{i s o b a r i c} + W_{i s o c h o r i c}$
$\Rightarrow 112.5 + 700 + 0$
$813 J$

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Similar questions

Q. A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure

P_{i} = 10^{5} P a

and volume

V_{i} = 10^{- 3} P a

changes to a final state at

P_{f} = (\frac{1}{32}) \times 10^{5} P a

and

V_{f} = 8 \times 10^{- 3} m^{3}

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P^{3} V^{5} =

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followed by an isochoric (isovolumetric) process at volume

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Q. A gas is enclosed in a cylinder with a movable friction less piston. Its initial thermodynamic state at pressure

P_{i} = 10^{5} P a

and volume

V_{i} = 10^{- 3} m^{3}

changes to a final state at

P_{f} = (1 / 32) \times 10^{5} P a

and

V_{f} = 8 \times 10^{- 3} m^{3}

in an adiabatic quasi-static process, such that

P^{3} V^{5}

=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

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Q. A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure

P_{i} = 10^{5} P a

and volume

V_{i} = 10^{- 3} P a

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P_{f} = (\frac{1}{32}) \times 10^{5} P a

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V_{f} = 8 \times 10^{- 3} m^{3}

in an adiabatic quasi-static process, such that

P^{3} V^{5} =

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P_{i}

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P_{i}, V_{i}

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V_{f}

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The correct option is D 813 Jfor adiabatic processP3V3=CPV5/3=Kr=5/3wadiabatic=11−r(P2V2−P1V1)=11−53(132×105×8103−105×1103)=112.5 JWisobaric=PΔV=105×(8−1)×10−3=7×100=700 J⇒Wadiabatic+Wisobaric+Wisochoric⇒112.5+700+0 813 J