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Summary STUDY MATERIAL FOR JEE ASPIRANTS

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This study includes all the thing an JEE student need to study . The way he wants to crack this examination will need overview of important formula of each chapter which we have provided in this document. All the best for jee preparation and make sure that you can crack it for your parents , your teacher and your mentor . Best of luck

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RACE # 45 THER MODYNAM ICS AND THERMOCHEMISTRY C HE M I S T R Y
FLOT
1. What is the difference between change in enthalpy and change in internal energy at constant volume :
(A) 0 (B) VdP (C) –VdP (D) PdV
2. The work done in an adiabatic process on an ideal gas by a constant external pressure would be :
(A) Zero (B) DE (C) DH (D) q
3. Which of the following is correct option for free expansion of an ideal gas under adiabatic condition ?
(A) q = 0, DT ¹ 0, w = 0 (B) q ¹ 0, DT = 0, w = 0
(C) q = 0, DT = 0, w = 0 (D) q = 0, DT < 0, w ¹ 0
4. Consider the cyclic process R ® S ® R as shown in the fig. You are told that one of the path is adiabatic and
the other one isothermal. Which one of the following is(are) true?
(A) Process R ® S is isothermal
(B) Process S ® R is adiabatic
(C) Process R ® S is adiabatic
(D) Such a graph is not possible
5. Calculate the heat needed to raise the temperature of 20 g iron from 25ºC to 500ºC, if specific heat capacity of
iron is 0.45 JK–1 g–1 .
(A) 4275 J (B) 225 J (C) 15.66 J (D) 2250 J
6. The ratio of slopes of P–V plots for reversible adiabatic process and reversible isothermal process of an ideal
gas is equal to :
(A) g (B) 1 – g (C) (g – 1) (D) 1/g
7. For an ideal gas having molar mass M, specific heat at constant pressure can be given as :
g RM R RM gR
(A) g –1
(B) M( g - 1)
(C) g –1
(D) M( g - 1)

8. P–V plots for two gases during an adiabatic process are given in the figure :




Volume

Plot A and plot B should correspond to : (Assume ideal behaviour)
(A) He and O2 (B) SO2 and Ar (C) O2 and He (D) Both (B) and (C)
9. An ideal gas with Cv = 3R expands adiabatically into a vacuum thus doubling its volume. The final temperatue
is given by :
(A) T2 = T1[2–1/3] (B) T2 = T1 (C) T2 = 2T1 (D) T2 = T1
2

3
10. A monoatomic ideal gas (CV = R) is allowed to expand adiabatically and reversibly from initial volume of 8
2
L at 300 K to a volume of V2 at 250 K. V2 is : (Given (4.8)1/2 = 2.2)
(A) 10.56 L (B) 17.6 L
(C) 11.52 L (D) Expansion not possible since temperature has decreased
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