Some gas (Cp / Cv=γ=1.25) follows the cycle A B C D A as shown in the figure. Find the ratio of the energy given by the gas to its surroundings during the isochoric section of the cycle to the work done in expansion of the gas during the isobaric section of the cycle.

Question

Some gas (Cp / Cv=&gamma;=1.25) follows the cycle A B C D A as shown in the figure. Find the ratio of the energy given by the gas to its surroundings during the isochoric section of the cycle to the work done in expansion of the gas during the isobaric section of the cycle. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

For isochoric process: d Q=n Cv d T For isobaric process: d W=n R d T=n(Cp-Cv) d T ∴ (d Q/d W)=(Cv/Cp-Cv)=(1/1.25-1)=4

Q. Some gas $(C_{p} / C_{v} = γ = 1.25)$ follows the cycle $A BC D A$ as shown in the figure. Find the ratio of the energy given by the gas to its surroundings during the isochoric section of the cycle to the work done in expansion of the gas during the isobaric section of the cycle.

For isochoric process:
$d Q = n C_{v} d T$
For isobaric process:
$d W = n R d T = n (C_{p} - C_{v}) d T$
$∴ \frac{d Q}{d W} = \frac{C _{v}}{C _{p} - C _{v}} = \frac{1}{1.25 - 1} = 4$

Q. Some gas (Cp​/Cv​=γ=1.25) follows the cycle ABCDA as shown in the figure. Find the ratio of the energy given by the gas to its surroundings during the isochoric section of the cycle to the work done in expansion of the gas during the isobaric section of the cycle.

For isochoric process: dQ=nCv​dT For isobaric process: dW=nRdT=n(Cp​−Cv​)dT ∴dWdQ​=Cp​−Cv​Cv​​=1.25−11​=4

Q. Some gas $(C_{p} / C_{v} = γ = 1.25)$ follows the cycle $A BC D A$ as shown in the figure. Find the ratio of the energy given by the gas to its surroundings during the isochoric section of the cycle to the work done in expansion of the gas during the isobaric section of the cycle.

For isochoric process:
$d Q = n C_{v} d T$
For isobaric process:
$d W = n R d T = n (C_{p} - C_{v}) d T$
$∴ \frac{d Q}{d W} = \frac{C _{v}}{C _{p} - C _{v}} = \frac{1}{1.25 - 1} = 4$