One mole of diatomic gas is taken through a cyclic process- Pressure- volume and temperature of gas at A is P0- V0 and T0- List-I gives some physical quantities and List-II gives their possible magnitudes-List-IList-II1 Work done in process BC P 32 P 0 V 02 Work done in process CA Q 2 P 0 V 03 Change in internal energy R3P0V0 during process BC4 Change in internal energy S 5 POV 0 during process CAT152P0V0U 0shown in figure- Then match List-I with List-II-A- 1- R -2- S -3- T -4- U B- 1-P- 2-R- 3-Q- 4-SC- 1-R- 2-U- 3-T- 4-T

The correct option is C (1) (R), (2) (U), (3) (T), (4) (T)Volume nbsp;at nbsp;point nbsp;A nbsp;= nbsp;V0Volume nbsp;at nbsp;point nbsp;B nbs ...

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Standard XII

Chemistry

Internal Energy

One mole of d...

Question

One mole of diatomic gas is taken through a cyclic process. Pressure, volume and temperature of gas at A is $P_{0}, V_{0}$ and $T_{0}$ . List-I gives some physical quantities and List-II gives their possible magnitudes.

List-I
List-II

(1) Work done in process BC
(P) $\frac{3}{2} P_{0} V_{0}$

(2) Work done in process CA
(Q) $2 P_{0} V_{0}$

(3) Change in internal energy
during process BC
(R) $3 P_{0} V_{0}$

(4) Change in internal energy
during process CA
(S) $5 P_{0} V_{0}$

(T) $\frac{15}{2} P_{0} V_{0}$

(U) 0

If cyclic process is given on P-T graph as shown in figure. Then match List-I with List-II.

(1)-(R), (2)-(S), (3)-(T), (4)-(U)

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(1)-(P), (2)-(R), (3)-(Q), (4)-(S)

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(1)-(R), (2)-(U), (3)-(T), (4)-(T)

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(1)-(R), (2)-(P), (3)-(T), (4)-(U)

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Solution

The correct option is C (1)-(R), (2)-(U), (3)-(T), (4)-(T)
$V o l u m e a t p o i n t A = V_{0} V o l u m e a t p o i n t B = \frac{V_{0}}{4} V o l u m e a t p o i n t C = V_{0} Δ W_{B C} = P Δ V = 4 P_{0} (V_{0} - \frac{V_{0}}{4}) = 3 P_{0} V_{0} Δ W_{C A} = 0 (S i n c e Δ V = 0) Δ U_{B C} = n C_{v} Δ T = \frac{5}{2} R (3 T_{0}) = \frac{15}{2} P_{0} V_{0} Δ U_{C A} = \frac{- 15}{2} P_{0} V_{0} (S i n c e Δ T = - 3 T_{0})$

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One mole of diatomic gas is taken through a cyclic process. Pressure, volume and temperature of gas at A is $P_{0}$ , $V_{0}$ and $T_{0}$ . List-I gives some physical quantities and List-II gives their possible magnitudes.

List-I

List-II

(1) Work done in process BC	(P) $\frac{3}{2} P_{0} V_{0}$
(2) Work done in process CA	(Q) $2 P_{0} V_{0}$
(3) Change in internal energy during process BC	(R) $3 P_{0} V_{0}$
(4) Change in internal energy during process CA	(S) $5 P_{0} V_{0}$
	(T) $\frac{15}{2} P_{0} V_{0}$
	(U) 0

If cyclic process is given on P-V graph as shown in figure then match List-I with List-II.

In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by $T Δ X$ , where $T$ is temperature of the system and $Δ X$ is the infinitesimal change in a thermodynamic quantity $X$ of the system. For a mole of monatomic ideal gas $X = \frac{3}{2} R ln (\frac{T}{T_{A}}) + Rln (\frac{V}{V_{A}}) .$ Here, $R$ is gas constant, $V$ is volume of gas, $T_{A}$ and $V_{A}$ are constants.

The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.

	List-I		List-II
(i)	Work done by the system in process $1 \to 2 \to 3$	(P)	$\frac{1}{3} {R T}_{0} ln 2$
(ii)	Change in intermal energy in process $1 \to 2 \to 3$	(Q)	$\frac{1}{3} {R T}_{0}$
(iii)	Heat absorbed by the system in process $1 \to 2 \to 3$	(R)	${R T}_{0}$
(iv)	Heat absotbed by the system in process $1 \to 2$	(S)	$\frac{4}{3} {R T}_{0}$
		(T)	$\frac{1}{3} {R T}_{0} (3 + ln 2)$
		(U)	$\frac{5}{6} {R T}_{0}$

If the process carricd out on one mole of monoatomic ideal gas is as shown in figure in the PV-diagram with $P_{0} V_{0} = \frac{1}{3} {R T}_{0}$ , the correct match is,

Q. One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the

P V

-diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-1 with the corresponding statements in List-II.

List-I	List-II
P. In process I	1. Work done by the gas is zero
Q. In process II	2 Temperature of the gas remains unchanged
R. In process III	3 No heat is exchanged between the gas and its surroundings
S. In process IV	4 Work done by the gas is $6 P_{0} V_{0}$

The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.

	List-I		List-II
(i)	Work done by the system in process $1 \to 2 \to 3$	(P)	$\frac{1}{3} {R T}_{0} ln 2$
(ii)	Change in intermal energy in process $1 \to 2 \to 3$	(Q)	$\frac{1}{3} {R T}_{0}$
(iii)	Heat absorbed by the system in process $1 \to 2 \to 3$	(R)	${R T}_{0}$
(iv)	Heat absotbed by the system in process $1 \to 2$	(S)	$\frac{4}{3} {R T}_{0}$
		(T)	$\frac{1}{3} {R T}_{0} (3 + ln 2)$
		(U)	$\frac{5}{6} {R T}_{0}$

If the process carricd out on one mole of monoatomic ideal gas is as shown in figure in the PV-diagram with $P_{0} V_{0} = \frac{1}{3} {R T}_{0}$ , the correct match is,

The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.

	List-I		List-II
(i)	Work done by the system in process $1 \to 2 \to 3$	(P)	$\frac{1}{3} {R T}_{0} ln 2$
(ii)	Change in intermal energy in process $1 \to 2 \to 3$	(Q)	$\frac{1}{3} {R T}_{0}$
(iii)	Heat absorbed by the system in process $1 \to 2 \to 3$	(R)	${R T}_{0}$
(iv)	Heat absotbed by the system in process $1 \to 2$	(S)	$\frac{4}{3} {R T}_{0}$
		(T)	$\frac{1}{3} {R T}_{0} (3 + ln 2)$
		(U)	$\frac{5}{6} {R T}_{0}$

If the process carricd out on one mole of monoatomic ideal gas is as shown in figure in the PV-diagram with $P_{0} V_{0} = \frac{1}{3} {R T}_{0}$ , the correct match is,

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The correct option is C (1)-(R), (2)-(U), (3)-(T), (4)-(T)Volume at point A = V0Volume at point B = V04Volume at point C = V0ΔWBC=PΔV = 4P0(V0− V04) = 3P0V0ΔWCA = 0 (Since ΔV = 0) ΔUBC= nCvΔT = 52R(3T0) =152P0V0ΔUCA=−152P0V0 (Since ΔT=−3T0)