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-IList-II1 Work done in process BC P 3-2 P 0 V 02 Work done in process CA Q 2 P 0 V 03 Change in internal energy during process BC4 Change in internal energy S 5 P 0 V 0 during process CA T 15-2 P 0 V 0cyclic process is given on P - V graph as shown in figure then match List-I with List-II-A- 1-P- 2-S- 3-Q- 4-TB- 1-Q- 2-P- 3-S- 4-TC- 1-R- 2-S- 3-S- 4-TD- 1-Q- 2-R- 3-S- 4-U

The correct option is B (1) (Q), (2) (P), (3) (S), (4) (T)WAC=P∆V nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp;=2P0V0For nbsp;CA, nbsp; ...

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Byju's Answer

Standard XII

Chemistry

Relation between P, V, T, Gamma in Adiabatic Proceses

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-V graph as shown in figure then match List-I with List-II.

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

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

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

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

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Solution

The correct option is B (1)-(Q), (2)-(P), (3)-(S), (4)-(T)
$W_{A C} = P Δ V = 2 P_{0} V_{0} F o r C A, A s s l o p e o f C A i s 1; P_{0} \frac{P_{0}}{V_{0}} = \frac{P}{V} \Rightarrow W_{C A} = \int P d V = \int_{0}^{V_{0}} \frac{P_{0} V}{V_{0}} d V W_{C A} = \frac{P_{0}}{2 V_{0}} {(V^{2}]}_{0}^{2} = \frac{- P_{0}}{2 V_{0}} 3 {V_{0}}_{0}^{2} = - \frac{3}{2} P_{0} V_{0} Δ U_{B C} = n C_{V} Δ T F o r d i a t o m i c g a s n C_{V} = \frac{5}{2} R = \frac{5}{2} R \times 2 T_{0} = 5 P_{0} V_{0} Δ U_{C A} = n C_{V} Δ T = - \frac{5}{2} R \times 3 T_{0} = \frac{- 15}{2} P_{0} V_{0}$

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Similar questions

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.

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 $6 P_{0} V_{0}$

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. List - II Gives the coordinates of a point P that divides the line segment AB joining the points given in List -I, match them correctly.

\begin{matrix} L i s t - I & L i s t I I (P) & A (- 1, 3) a n d B (- 5, 6) I n t e r n a l l y i n t h e r a t i o 1 : 2 & (1) & (7, 3) (Q) & A (- 2, 1) a n d B (1, 4) i n t e r n a l l y i n t h e r a t i o 2 : 1 & (2) & (0, 3) (R) & A (1, 7) a n d B (3, 4) i n t e r n a l l y i n t h e r a t i o 3 : 2 & (3) & (\frac{11}{5}, \frac{26}{5}) (S) & a (4, - 3) a n d B (8, 5) i n t e r n a l l y i n t h e r a t i o 3 : 1 & (4) & (- \frac{7}{3}, 4) \end{matrix}

Q. List-II gives distance between pair of points given in List-I, match them correctly.

\begin{matrix} L i s t - I & L i s t I I (P) & (- 5, 7), (- 1, 3) & (1) & 5 (Q) & (5, 6), (1, 3) & (2) & \sqrt{8} (R) & (\sqrt{3} + 1, 1), (0, \sqrt{3}) & (3) & \sqrt{6} (S) & (0, 0), (- \sqrt{3}, \sqrt{3}) & (4) & 4 \sqrt{2} \end{matrix}

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The correct option is B (1)-(Q), (2)-(P), (3)-(S), (4)-(T)WAC=PΔV =2P0V0For CA, As slope of CA is 1 ;P0P0V0=PV ⇒WCA=∫PdV =∫V02V0P0VV0dVWCA=P02V0(V2]V02V0 =−P02V03V02 =−32P0V0ΔUBC=nCVΔTFor diatomic gas nCV=52R =52R×2T0 =5P0V0ΔUCA=nCVΔT =−52R×3T0 =−152P0V0