One mole of a monatomic ideal gas is taken through a cycle $A B C D A$ as shown in the $P - V$ diagram. List 2 gives the characteristics involved in the cycle. Match them with each of the processes given in List 1.

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Solution

(A) It is an isobaric process.
Volume decreases, therefore temperature decreases.If temperature decreases, internal energy decreases.
From the first law of thermodynamics,
$Q = U + W$
Since work done is negative, Heat is lost . It is the work done on the gas.
(B) It is an isochoric process.
Pressure decreases, therefore temperature decreases and hence internal energy decreases.Work done is always zero. Heat is lost by the system.
(C) It is an isobaric process.Volume increases, therefore temperature increases and hence internal energy will increase. Heat is gained by the system since it is positive in sign. Work is done by the gas.
(D) $T_{D}$ = $\frac{P (9 V)}{n R}$ = $\frac{9 P V}{n R}$
$T_{A}$ = $\frac{(3 P) (3 V)}{n R}$ = $\frac{9 P V}{n R}$ . So it is an isothermal process.
$W = n R T l n \frac{V_{A}}{V_{D}}$
Work done will be negative and hence heat is lost. It is the work done on the gas.

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\begin{matrix} Column I & Column II A) Process A \to B & p) Internal energy decreases. B) Process B \to C & q) Internal energy increases. C) Process C \to D & r) Heat is lost D) Process D \to A & s) Heat is gained. t) Work is done one the gas. \end{matrix}

A B C D A

P - V

A B C D A

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