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1

IIT-JEE 2011 Paper 2 Offline

MCQ (Single Correct Answer)

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$$ \to $$ B (P) Internal energy decreases.
(B) Process B$$ \to $$C (Q) Internal energy increase.
(C) Process C$$ \to $$D (R) Heat is lost.
(D) Process D$$ \to $$A (S) Heat is gained.
(T) Work is done on the gas.

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

Explanation

Process A $$\to$$ B,

It is a isobaric process

P = constant, V $$\propto$$ T

$$\because$$ VB < VA $$\Rightarrow$$ TB < TA

$$\Delta$$U = nCV$$\Delta$$T = $$-$$ve

Hence internal energy decreases.

$$\Delta$$Q = nCP$$\Delta$$T = $$-$$ve

Hence heat is lost.

$$\Delta$$W = nR$$\Delta$$T = $$-$$ve

Hence, work is done on the gas.

Process, B $$\to$$ C,

It is a isochoric process.

V = constant, P $$\propto$$ T

$$\because$$ PC < PB $$\Rightarrow$$ TC < TB

$$\Delta$$U = nCV$$\Delta$$T = $$-$$ve

Hence, internal energy decreases.

$$\Delta$$W = 0, $$\Delta$$Q = $$\Delta$$U = $$-$$ve

Hence, heat is lost.

Process C $$\to$$ D,

It is a isobaric process.

P = constant, V $$\propto$$ T

$$\because$$ VD > VC $$\Rightarrow$$ TD > TC

$$\Delta$$U = nCV$$\Delta$$T = +ve

Hence internal energy increases.

$$\Delta$$Q = nCP$$\Delta$$T = +ve

Hence, heat is gained.

$$\Delta$$W = nR$$\Delta$$T = +ve

Hence, work is done by the gas.

Process, D $$\to$$ 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.

$$\therefore$$ $$\Delta$$U = 0

$$\because$$ VA < VD

$$\Delta$$W = $$-$$ve, hence work done is done on the gas.

$$\Delta$$Q = $$\Delta$$W = $$-$$ve, hence heat is lost.

2

IIT-JEE 2011 Paper 1 Offline

MCQ (Single Correct Answer)
5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be T1, the work done in the process is
A
$${9 \over 8}R{T_1}$$
B
$${3 \over 2}R{T_1}$$
C
$${15 \over 8}R{T_1}$$
D
$${9 \over 2}R{T_1}$$

Explanation

Initially

V1 = 5.6 l, T1 = 273 K, P1 = 1 atm,

$$\gamma = {5 \over 3}$$ (For monoatomic gas)

The number of moles of gas is $$n = {{5.6l} \over {22.4l}} = {1 \over 4}$$

Finally (after adiabatic compression)

V2 = 0.7 l

For adiabatic compression $${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$$

$$\therefore$$ $${T_2} = {T_1}{\left( {{{{V_1}} \over {{V_2}}}} \right)^{\gamma - 1}} = {T_1}{\left( {{{5.6} \over {0.7}}} \right)^{{5 \over 3} - 1}} = {T_1}{(8)^{2/3}} = 4{T_1}$$

Work done during an adiabatic process is

$$W = {{nR[{T_1} - {T_2}]} \over {(\gamma - 1)}} = {{{1 \over 4}R[{T_1} - 4{T_1}]} \over {\left[ {{5 \over 3} - 1} \right]}} = - {9 \over 8}R{T_1}$$

Negative sign shows that work is done on the gas.

3

IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)
A real gas behaves like an ideal gas if its
A
pressure and temperature are both high
B
pressure and temperature are both low
C
pressure is high and temperature is low
D
pressure is low and temperature is high

Explanation

In an ideal gas, the average force of attraction between the molecules and volume of the molecules (in comparison to volume of the gas) are negligibly small. These conditions are satisfied for a real gas when pressure is low and temperature is high.

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