Question

A piston-cylinder device operates 1 kg of fluid at 20 atm. Pressure. The initial volume is 0.04 m³. The fluid is allowed to expand reversibly following a process pV¹45= constant so that volume becomes double. The fluid is the cooled at constant pressure until the piston comes back to the original position. Keeping the piston unaltered, heat is added reversibly to restore it to the initial pressure. (i) Calculate the work done in the cycle and (ii) calculate the work done in only expansion process.​

Answer 1

Explanation:

a

3

b−a

2

b+5ab−5b

By taking a

2

b common in the first two term and 5b as common in the second two term

a

3

b−a

2

b+5ab−5b=a

2

b(a−1)+5ab(a−1)

So we get,

a

3

b−a

2

b+5ab−5b=(a−1)(a

2

b+5b)

By taking b as common. We get,

a

3

b−a

2

b+5ab−5b=b(a−1)(a

2

+5

Answer 2

Answer:

The work done in complete cycle is 18.6kJ and in only expansion is 48.27J.

Explanation:

Given the operating pressure on fluid by piston-cylinder device,

$P_i= 20 atm.$

Initial volume of the fluid, $V_i= 0.04 m^3$

Volume gets doubled, therefore $V_f= 2\times 0.04 = 0.08 m^3$

Fluid is the cooled at constant pressure, so reaches initial volume again.

Then heat is added to revert back to the initial pressure, $P_i$.

As shown in the figure, a complete cycle goes from point 1 to 2 and to 3.

Given the expansion follows the polytropic process, and the relation is

$PV^{\gamma}= constant\implies PV^{1.45}= constant$

$\implies P_iV_i^{1.45}= P_fV_f^{1.45}$

$\implies 20\times 0.04 ^{1.45}= P_f\times 0.08^{1.45}$

$\implies P_f= \dfrac{20\times 0.04 ^{1.45}}{0.08^{1.45}} =7.32atm$

The work done in polytropic process is

$W= \dfrac{P_iV_i-P_fV_f}{\gamma-1}$

(i) Work done in complete cycle is from point 1 to 2 to 3 to 1.

$W_{1231}= W_{12}+W_{23}+W_{31}$

Substituting the values from point 1 to 2 by converting the pressure units from atm to kpa, the work done is

$W_{12}= \dfrac{20\times101.325\times 0.04 -7.32\times101.325\times0.08}{1.45-1}$

      $= \dfrac{81.06 -59.33}{0.45}= \dfrac{21.72}{0.45}=48.27kJ$

From point 2 to 3, the pressure is $P_f$ and is constant, and the volume changes from $V_f$ to $V_i$.

Therefore, the work done from point 2 to 3 is

$W_{23}= P_2\big(V_i-V_f\big)$

      $=7.32\times101.325\big(0.04-0.08\big)$

      $=741.699\big(-0.04\big)=-29.67kJ$

From point 3 to 1, the volume is $V_i$ and is constant, so change in volume $dV=0$.

Therefore, the work done from point 3 to 1 is given by

$W_{31}= PdV=0$

Thus, the total work done in complete cycle is

$W_{1231}= 48.27-29.67+0=18.6kJ$

(ii) The work done in only the expansion process is from point 1 to 2 is

      $W_{12}=48.27J$

Hence, the work done in complete cycle is 18.6kJ and in only expansion is 48.27J.

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