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1

AIEEE 2009

MCQ (Single Correct Answer)
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
$$W = {W_{AB}} + {W_{BC}} + {W_{CD}} + {W_{DA}}$$
$$ = 400R + 2.303nRT\log {{{P_B}} \over {{P_C}}} + \left( { - 400R} \right) - 414R$$
$$ = 2.303 \times 2R \times 500\log {{2 \times {{10}^5}} \over {1 \times {{10}^5}}} - 414R$$
$$ = 693.2\,R - 414\,R = 279.2\,R$$
2

AIEEE 2009

MCQ (Single Correct Answer)
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 $$ + \,414\,R.$$
3

AIEEE 2009

MCQ (Single Correct Answer)
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
$$W = nR\left( {{T_2} - {T_1}} \right) = 2R\left( {500 - 300} \right) = 400R$$
4

AIEEE 2009

MCQ (Single Correct Answer)
A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature $$\theta $$ along the length $$x$$ of the bar from its hot end is best described by which of the following figures?
A
B
C
D

Explanation

The heat flow rate is given by

$${{dQ} \over {dt}} = {{kA\left( {{\theta _1} - \theta } \right)} \over x}$$

$$ \Rightarrow {\theta _1} - \theta $$ $$ = {x \over {kA}}{{dQ} \over {dt}} $$

$$\Rightarrow \theta $$ $$ = {\theta _1} - {x \over {kA}}{{dQ} \over {dt}}$$

where $${\theta _1}$$ is the temperature of hot end and $$\theta $$ is temperature at a distance $$x$$ from hot end.
The above equation can be graphically represented by option $$(a).$$

Questions Asked from Heat and Thermodynamics

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