A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at pressure P0, volume V0, and temperature T0. If the gas mixture is adiabatically compressed to a volume V_0 \over 4 , then the correct statement(s) is/are (Given, 21.2 = 2.3; 23.2 = 9.2; R is a gas constant)

The final pressure of the gas mixture after compression is in between 9P0 and 10P0, Adiabatic constant of the gas mixture is 1.6, The work |W| done during the process is 13RT0

The final pressure of the gas mixture after compression is in between 9P0 and 10P0

The average kinetic energy of the gas mixture after compression is in between 18RT0 and 19RT0

Adiabatic constant of the gas mixture is 1.6

The work |W| done during the process is 13RT0

Heat and Thermodynamics PYQ — JEE Advanced 2019 · Shift 2 · Q44 with Solution

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Correct answer: A, C, D

Find heat capacities of the mixture

For ideal gases:

Monatomic gas: $C_V = \dfrac{3R}{2}$ , $C_P = \dfrac{5R}{2}$
Rigid diatomic gas: $C_V = \dfrac{5R}{2}$ , $C_P = \dfrac{7R}{2}$

Given mixture:

$5$ moles monatomic
$1$ mole rigid diatomic

So total heat capacity at constant volume:

C_{V,\text{mix}} = 5\left(\frac{3R}{2}\right) + 1\left(\frac{5R}{2}\right) = \frac{15R}{2} + \frac{5R}{2} = 10R

Total heat capacity at constant pressure:

C_{P,\text{mix}} = 5\left(\frac{5R}{2}\right) + 1\left(\frac{7R}{2}\right) = \frac{25R}{2} + \frac{7R}{2} = 16R

Hence adiabatic constant:

\gamma = \frac{C_P}{C_V} = \frac{16R}{10R} = 1.6

So Option C is correct.

Use adiabatic relation to find final pressure

For an adiabatic process:

PV^\gamma = \text{constant}

Thus,

P_f = P_0\left(\frac{V_0}{V_f}\right)^\gamma

Given:

V_f = \frac{V_0}{4}

So,

P_f = P_0 (4)^{1.6}

Now,

4^{1.6} = (2^2)^{1.6} = 2^{3.2}

Given $2^{3.2} = 9.2$ .

Therefore,

P_f = 9.2 P_0

which lies between $9P_0$ and $10P_0$ .

So Option A is correct.

Find final temperature

For an adiabatic process of an ideal gas:

TV^{\gamma-1} = \text{constant}

Hence,

T_f = T_0\left(\frac{V_0}{V_f}\right)^{\gamma-1}

Here,

\gamma - 1 = 1.6 - 1 = 0.6

So,

T_f = T_0 \cdot 4^{0.6}

Now,

4^{0.6} = (2^2)^{0.6} = 2^{1.2}

Given $2^{1.2} = 2.3$ .

Thus,

T_f = 2.3T_0

Check average kinetic energy of the gas mixture

For an ideal gas mixture, total kinetic/internal energy is

U = C_V T = 10RT

for the whole mixture.

At final state,

U_f = 10R(2.3T_0) = 23RT_0

If the option means average kinetic energy of the whole mixture, it is certainly not between $18RT_0$ and $19RT_0$ .

Even if one interprets “average kinetic energy” per mole or per molecule, the value is proportional to $T_f = 2.3T_0$ , not to a number in the range $18RT_0$ to $19RT_0$ .

So Option B is incorrect.

Find work done in adiabatic compression

Since the process is adiabatic:

Q=0

Using first law,

\Delta U = Q - W = -W

where $W$ is work done by the gas.

Therefore,

|W| = \Delta U = C_V (T_f - T_0)

Now,

|W| = 10R(2.3T_0 - T_0) = 10R(1.3T_0) = 13RT_0

So Option D is correct.

Final conclusion

Correct options are:

\boxed{A,\ C,\ D}

This matches the stored correct answer.

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View written solutionFree

Correct answer: A, C, D

Find heat capacities of the mixture

For ideal gases:

Monatomic gas: $C_V = \dfrac{3R}{2}$ , $C_P = \dfrac{5R}{2}$
Rigid diatomic gas: $C_V = \dfrac{5R}{2}$ , $C_P = \dfrac{7R}{2}$

Given mixture:

$5$ moles monatomic
$1$ mole rigid diatomic

So total heat capacity at constant volume:

C_{V,\text{mix}} = 5\left(\frac{3R}{2}\right) + 1\left(\frac{5R}{2}\right) = \frac{15R}{2} + \frac{5R}{2} = 10R

Total heat capacity at constant pressure:

C_{P,\text{mix}} = 5\left(\frac{5R}{2}\right) + 1\left(\frac{7R}{2}\right) = \frac{25R}{2} + \frac{7R}{2} = 16R

Hence adiabatic constant:

\gamma = \frac{C_P}{C_V} = \frac{16R}{10R} = 1.6

So Option C is correct.

Use adiabatic relation to find final pressure

For an adiabatic process:

PV^\gamma = \text{constant}

Thus,

P_f = P_0\left(\frac{V_0}{V_f}\right)^\gamma

Given:

V_f = \frac{V_0}{4}

So,

P_f = P_0 (4)^{1.6}

Now,

4^{1.6} = (2^2)^{1.6} = 2^{3.2}

Given $2^{3.2} = 9.2$ .

Therefore,

P_f = 9.2 P_0

which lies between $9P_0$ and $10P_0$ .

So Option A is correct.

Find final temperature

For an adiabatic process of an ideal gas:

TV^{\gamma-1} = \text{constant}

Hence,

T_f = T_0\left(\frac{V_0}{V_f}\right)^{\gamma-1}

Here,

\gamma - 1 = 1.6 - 1 = 0.6

So,

T_f = T_0 \cdot 4^{0.6}

Now,

4^{0.6} = (2^2)^{0.6} = 2^{1.2}

Given $2^{1.2} = 2.3$ .

Thus,

T_f = 2.3T_0

Check average kinetic energy of the gas mixture

For an ideal gas mixture, total kinetic/internal energy is

U = C_V T = 10RT

for the whole mixture.

At final state,

U_f = 10R(2.3T_0) = 23RT_0

If the option means average kinetic energy of the whole mixture, it is certainly not between $18RT_0$ and $19RT_0$ .

Even if one interprets “average kinetic energy” per mole or per molecule, the value is proportional to $T_f = 2.3T_0$ , not to a number in the range $18RT_0$ to $19RT_0$ .

So Option B is incorrect.

Find work done in adiabatic compression

Since the process is adiabatic:

Q=0

Using first law,

\Delta U = Q - W = -W

where $W$ is work done by the gas.

Therefore,

|W| = \Delta U = C_V (T_f - T_0)

Now,

|W| = 10R(2.3T_0 - T_0) = 10R(1.3T_0) = 13RT_0

So Option D is correct.

Final conclusion

Correct options are:

\boxed{A,\ C,\ D}

This matches the stored correct answer.

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