### JEE Mains Previous Years Questions with Solutions

4.5
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1

### AIEEE 2010

A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of the gas increases from $V$ to $32$ $V$, the efficiency of the engine is
A
$0.5$
B
$0.75$
C
$0.99$
D
$0.25$

## Explanation

${T_1}{V^{\gamma - 1}} = {T_2}{\left( {32V} \right)^{\gamma - 1}}$
$\Rightarrow {T_1} = {\left( {32} \right)^{\gamma - 1}}.{T_2}$
For diatomic gas, $\gamma = {7 \over 5}$
$\therefore$ $\gamma - 1 = {2 \over 5}$
$\therefore$ ${T_1} = {\left( {32} \right)^{{2 \over 5}}}.{T_2} \Rightarrow {T_1} = 4{T_2}$
Now, efficiency $= 1 - {{{T_2}} \over {{T_1}}}$
$= 1 - {{{T_2}} \over {4{T_2}}}$
$= 1 - {1 \over 4}$
$= {3 \over 4}$
$= 0.75.$
2

### AIEEE 2009

Two moles of helium gas are taken over the cycle $ABCD$, as shown in the $P$-$T$ diagram.

The net work done on the gas in the cycle $ABCDA$ is:

A
$276$ $R$
B
$1076$ $R$
C
$1904$ $R$
D
zero

## Explanation

The net work in the cycle $ABCD$ is
$W = {W_{AB}} + {W_{BC}} + {W_{CD}} + {W_{DA}}$
$= 400R + 2.303nRT\log {{{P_B}} \over {{P_C}}} + \left( { - 400R} \right) - 414R$
$= 2.303 \times 2R \times 500\log {{2 \times {{10}^5}} \over {1 \times {{10}^5}}} - 414R$
$= 693.2\,R - 414\,R = 279.2\,R$
3

### AIEEE 2009

Two moles of helium gas are taken over the cycle $ABCD,$ as shown in the $P$-$T$ diagram.

The work done on the gas in taking it from $D$ to $A$ is :

A
$+414$ $R$
B
$-690$ $R$
C
$+690$ $R$
D
$-414$ $R$

## Explanation

Work done by the system in the isothermal process
$DA$ is $W = 2.303nRT\,{\log _{10}}{{{P_D}} \over {{P_A}}}$
$= 2.303 \times 2R \times 300{\log _{10}}{{1 \times {{10}^5}} \over {2 \times {{10}^5}}} = - 414R.$
Therefore work done on the gas is $+ \,414\,R.$
4

### AIEEE 2009

Two moles of helium gas are taken over the cycle $ABCD,$ as shown in the $P$-$T$ diagram.

Assuming the gas to be ideal the work done on the gas in taking it from $A$ to $B$ is :

A
$300$ $R$
B
$400$ $R$
C
$500$ $R$
D
$200$ $R$

## Explanation

$A$ to $B$ is an isobaric process. The work done
$W = nR\left( {{T_2} - {T_1}} \right) = 2R\left( {500 - 300} \right) = 400R$