Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

Numerical

The average translational kinetic energy of N_{2} gas molecules at .............$$^\circ$$C becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt. (Given k_{B} = 1.38 $$\times$$ 10^{$$-$$23} J/K) (Fill the nearest integer).

Your Input ________

Correct Answer is **500**

Given, the average translational kinetic energy of dinitrogen (N_{2}) = Kinetic energy of an electron .... (i)

Translational kinetic energy of dinitrogen (N_{2})

$$KE = {3 \over 2}{K_B}T$$

Here, T = temperature of the gas,

and K_{B} = Boltzmann constant.

Kinetic energy of an electron = eV

Given, the potential differential of an electron, V = 0.1 V

Substituting the values in the Eq. (i), we get

$${3 \over 2}{K_B}T = eV$$

$$ \Rightarrow {3 \over 2} \times 1.38 \times {10^{ - 23}} \times T = 1.6 \times {10^{ - 19}} \times (0.1)$$

$$T = 773K = 773 - 273^\circ C = 500^\circ C$$

Translational kinetic energy of dinitrogen (N

$$KE = {3 \over 2}{K_B}T$$

Here, T = temperature of the gas,

and K

Kinetic energy of an electron = eV

Given, the potential differential of an electron, V = 0.1 V

Substituting the values in the Eq. (i), we get

$${3 \over 2}{K_B}T = eV$$

$$ \Rightarrow {3 \over 2} \times 1.38 \times {10^{ - 23}} \times T = 1.6 \times {10^{ - 19}} \times (0.1)$$

$$T = 773K = 773 - 273^\circ C = 500^\circ C$$

2

Numerical

The temperature of 3.00 mol of an ideal diatomic gas is increased by 40.0$$^\circ$$C without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of workdone by the gas is $${x \over {10}}$$. Then the value of x (round off to the nearest integer) is ___________. (Given R = 8.31 J mol^{$$-$$1} K^{$$-$$1})

Your Input ________

Correct Answer is **25**

Given, the number of diatomic moles, n = 3 mol

The increase in temperature of the diatomic mole,

$$\Delta$$T = 40$$^\circ$$C

Now, the degree of freedom

f = linear + rotational + no oscillation

f = 3 + 2 + 0 $$\Rightarrow$$ f = 5

Change in internal energy,

$$\Delta$$U = nC_{v}$$\Delta$$T .... (i)

where, $${C_v} = {f \over 2}R = {5 \over 2}R$$

Substituting the value in Eq. (i), we get

$$\Delta U = {{5R} \over 2}nR\Delta T$$

Now, work done by the gas for isobaric process,

$$W = p\Delta V = nR\Delta T$$

The ratio of the change in internal energy to the work done by the gas,

$${{\Delta U} \over W} = {{{5 \over 2}nR\Delta T} \over {nR\Delta T}}$$

$$ = {{\Delta U} \over W} = {5 \over 2}$$

Multiply and divide the above equation with 5, we get

$${{\Delta U} \over W} = {{5 \times 5} \over {2 \times 5}} = {{25} \over {10}}$$

Comparing with given equation, $${{\Delta U} \over W} = {x \over {10}}$$

The value of the x = 25.

The increase in temperature of the diatomic mole,

$$\Delta$$T = 40$$^\circ$$C

Now, the degree of freedom

f = linear + rotational + no oscillation

f = 3 + 2 + 0 $$\Rightarrow$$ f = 5

Change in internal energy,

$$\Delta$$U = nC

where, $${C_v} = {f \over 2}R = {5 \over 2}R$$

Substituting the value in Eq. (i), we get

$$\Delta U = {{5R} \over 2}nR\Delta T$$

Now, work done by the gas for isobaric process,

$$W = p\Delta V = nR\Delta T$$

The ratio of the change in internal energy to the work done by the gas,

$${{\Delta U} \over W} = {{{5 \over 2}nR\Delta T} \over {nR\Delta T}}$$

$$ = {{\Delta U} \over W} = {5 \over 2}$$

Multiply and divide the above equation with 5, we get

$${{\Delta U} \over W} = {{5 \times 5} \over {2 \times 5}} = {{25} \over {10}}$$

Comparing with given equation, $${{\Delta U} \over W} = {x \over {10}}$$

The value of the x = 25.

3

Numerical

A sample of gas with $$\gamma$$ = 1.5 is taken through an adiabatic process in which the volume is compressed from 1200 cm^{3} to 300 cm^{3}. If the initial pressure is 200 kPa. The absolute value of the workdone by the gas in the process = _____________ J.

Your Input ________

Correct Answer is **480**

v = 1.5

p_{1}v_{1}^{v} = p_{2}v_{2}^{v}

(200) (1200)^{1.5} = P^{2} (300)^{1.5}

P_{2} = 200 [4]^{3/2} = 1600 kPa

$$\left| {W.D.} \right| = {{{p_2}{v_2} - {p_1}{v_1}} \over {v - 1}} = \left( {{{480 - 240} \over {0.5}}} \right) = 480$$ J

p

(200) (1200)

P

$$\left| {W.D.} \right| = {{{p_2}{v_2} - {p_1}{v_1}} \over {v - 1}} = \left( {{{480 - 240} \over {0.5}}} \right) = 480$$ J

4

Numerical

A heat engine operates between a cold reservoir at temperature T_{2} = 400 K and a hot reservoir at temperature T_{1}. It takes 300 J of heat from the hot reservoir and delivers 240 J of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be ______________ K.

Your Input ________

Correct Answer is **500**

Q_{in} = 300 J ; Q_{out} = 240 J

Work done = Q_{in} $$-$$ Q_{out} = 300 $$-$$ 240 = 60 J

Efficiency = $${W \over {{Q_{in}}}} = {{60} \over {300}} = {1 \over 5}$$

efficiency = $$1 - {{{T_2}} \over {{T_1}}}$$

$${1 \over 5} = 1 - {{400} \over {{T_1}}} \Rightarrow {{400} \over {{T_1}}} = {4 \over 5}$$

T_{1} = 500 k

Work done = Q

Efficiency = $${W \over {{Q_{in}}}} = {{60} \over {300}} = {1 \over 5}$$

efficiency = $$1 - {{{T_2}} \over {{T_1}}}$$

$${1 \over 5} = 1 - {{400} \over {{T_1}}} \Rightarrow {{400} \over {{T_1}}} = {4 \over 5}$$

T

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