Question 153: Thermodynamics & KTG

Question

One mole of a monatomic gas is taken through a cycle ABCDA as shown in the PV diagram. Column II give the characteristics involved in the cycle. Match them with each of the processes given in Column I.

Column I

Column II

(A)
Process A$$ 	o $$ B
(P)
Internal energy decreases.

(B)
Process B$$ 	o $$C
(Q)
Internal energy increase.

(C)
Process C$$ 	o $$D
(R)
Heat is lost.

(D)
Process D$$ 	o $$A
(S)
Heat is gained.

(T)
Work is done on the gas.

Accepted Answer

Process A $$	o$$ B,
It is a isobaric process
P = constant, V $$\propto$$ T
$$\because$$ V
B
 V
C
$$\Rightarrow$$ T
D
> T
C
$$\Delta$$U = nC
V
$$\Delta$$T = +ve
Hence internal energy increases.
$$\Delta$$Q = nC
P
$$\Delta$$T = +ve
Hence, heat is gained.
$$\Delta$$W = nR$$\Delta$$T = +ve
Hence, work is done by the gas.
Process, D $$	o$$ A
According to ideal gas equation
$${{{P_A}{V_A}} \over {{T_A}}} = {{{P_D}{V_D}} \over {{T_D}}} \Rightarrow {{(3P)(3V)} \over {{T_A}}} = {{(P)(9V)} \over {{T_D}}} \Rightarrow {T_D} = {T_A}$$
Hence, it is a isothermal process.
$$	herefore$$ $$\Delta$$U = 0
$$\because$$ V
A
< V
D
$$\Delta$$W = $$-$$ve, hence work done is done on the gas.
$$\Delta$$Q = $$\Delta$$W = $$-$$ve, hence heat is lost.

Answer

(A)$$	o$$(P), (R), (T); (B)$$	o$$(P), R; (C)$$	o$$(Q), (S); (D)$$	o$$(R), (T)

Answer

(A)$$	o$$(P), (T); (B)$$	o$$(P), (R); (C)$$	o$$(Q), (S); (D)$$	o$$(R)

Answer

(A)$$	o$$(R), (T); (B)$$	o$$(P), (R); (C)$$	o$$(S); (D)$$	o$$(R), (T)

Answer

(A)$$	o$$(P), (R), (T); (B)$$	o$$(P), (R); (C)$$	o$$(Q); (D)$$	o$$(R), (T)

Thermodynamics & KTG

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