Option 2 : 2667 kJ/kg

__Concept:__

From the conservation of energy, Energy at input of mixing pipe = Energy at output of the mixing pipe

Energy carried by a flowing steam = \(mass\;flow\;rate \times \frac{{enthalpy}}{{unit\;mass}}\)

__Calculation:__

Energy of the stream 1 = ṁ_{1} × h_{1} = 1 × 3000 = 3000 kW

Energy of the stream 2 = ṁ_{2} × h_{2} = 2 × 2500 = 5000 kW

After the two streams mix into the one the resulting mass flow rate ṁ_{f }= 1 + 2 = 3 kg/s

Now the energy of the resulting mass = h_{f} × ṁ_{f}

From energy conservation E_{f }= E_{1} + E_{2}

⇒ h_{f} × 3 = 3000 + 5000

⇒ \({h_f} = \frac{{8000}}{3} = 2667 \;kJ/kg\)

A mixture of ideal gases has the following composition by mass:

N |
O |
CO |

60% |
30% |
10% |

If the Universal gas constant is 8314 J/kmol-K, the characteristic gas constant of the mixture (in J/kg.K) is _________.

**Concept:**

M_{T }R_{m} = m_{1}R_{1} + m_{2}R_{2 }+ m_{3}R_{3 }+ ..............+ m_{n}R_{n}

Rm = (m1R1 + m2R2 + m3R3 + ..............+ mnRn) / M_{T} ...................(1)

M_{T} = m1 + m2 + m_{3 }+...................+ m_{n}

R_{1, }R_{2}, R_{3 }are the characteristic gas constant of individual gases.

R_{m }= Characteristic gas constant of the mixture of gases

R (Characteristic gas constant) = (universal gas constant) / (molecular mass of gas)

R = R_{u }/ M

**Calculation:**

Given:

1,2, and 3 represent the gases N2, O2, and CO2 respectively.

m_{1 }= 0.6M_{T}, m_{2 }= 0.3M_{T},m_{3 }= 0.1M_{T}

R_{1 }= R_{u }/ 28, R_{2} = Ru / 32, R_{3 }= Ru / 44

R_{u} = 8314 J/kmol-k

Using equation 1,

\({R_m} = ({m_1}{R_1} + {m_2}{R_2} + {m_3}{R_3})/{M_T}\;\)

\({R_m} = \frac{{\left( {\frac{{0.6{M_T}{R_u}}}{{28}} + \frac{{0.3{M_T}{R_u}}}{{32}} + \frac{{0.1{M_T}{R_u}}}{{44}}} \right)}}{{{M_T}}}\)

\({R_m} = \frac{{0.6{R_u}}}{{28}} + \frac{{0.3{R_u}}}{{32}} + \frac{{0.1{R_u}}}{{44}}\)

R_{m} = 274.9961 J/kg-K

Option 1 : 1200 kPa

**Concept:**

As the process is adiabatic so applying adiabatic equation, i.e.,

\({P_1}V_1^γ = {P_2}V_2^γ\)

where, P = pressure, V = volume

**Calculation:**

**Given:**

P_{1} = 150 kPa, V_{1} = 1600 cm^{3}, V_{2} = 400 cm^{3}, γ = 1.5

⇒ \({P_2} = {P_1}{\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^γ } = 150{\left( {\frac{{1600}}{{400}}} \right)^{1.5}}\)

= 150 × 4Consider the following statements about the properties of perfect non reacting gas mixtures:

1) The total volume of a mixture is the sum of partial volumes at the same pressure and temperature.

2) The entropy of a mixture of gases is the sum of the the entropies of the constituents.

3) The total pressure of a mixture of gases is the sum of the partial pressures of the substances.

4) The mole fraction of a mixture of gases is equal to both pressure and volume fraction.

Which of the above statements is/are correct?

Option 1 : 1, 2, 3 and 4

**Explanation:**

**1. The partial volume of component i ^{th} in gaseous mixture is given by-**

V_{i} = \(\frac{{{{\rm{n}}_{\rm{i}}}{\rm{\bar RT}}}}{{\rm{P}}}\)

؞ Total volume of the mixture

V = \(\sum {{\rm{V}}_{\rm{i}}}\)

**2. The entropy of mixture-**

nS̅ = \(\sum {{\rm{n}}_{\rm{i}}}{{\rm{\bar S}}_{\rm{i}}}\)

= x_{1}S̅_{1} + x_{2}S̅_{2} +……….

**3. According to Dalton’s pressure law, the total pressure of a mixture of gases,**

P = P_{1} + P_{2} + P_{3} +……..

P = \(\sum {{\rm{P}}_{\rm{i}}}\)

**4. The pressure fraction**

So \(\frac{{{{\rm{P}}_{\rm{i}}}}}{{\rm{P}}}\) = \(\frac{{\left( {\frac{{{{\rm{n}}_{\rm{i}}}{\rm{\bar RT}}}}{{\rm{V}}}} \right)}}{{\left( {\frac{{{\rm{n\bar RT}}}}{{\rm{V}}}} \right)}}\)

Volume fraction

\(\frac{{{{\rm{V}}_{\rm{i}}}}}{{\rm{V}}}\) = \(\frac{{\left( {\frac{{{{\rm{n}}_{\rm{i}}}{\rm{\bar RT}}}}{{\rm{P}}}} \right)}}{{\left( {\frac{{{\rm{n\bar RT}}}}{{\rm{P}}}} \right)}}\)

**Hence all of the statements 1,2,3 and 4 are correct. So, option (1) is correct.**

Option 3 : \({C_p} - {C_v}\)

**Concept:**

Ideal-gas equation is given as, PV = nRoT

The characteristic gas equation is given as, PV = mRT

where R is known as Characteristic Gas Constant (or simply Gas Constant) and Ro is Universal Gas Constant.

\(R = \frac{{{R_0}}}{M}\)where M is molecular weight.

**And Characteristic gas constant is written as, **\(R={C_p} - {C_v}\)

where, C_{p} = heat capacity at constant pressure, C_{v} = heat capacity at constant volume

Option 1 : Chromatography

The correct answer is __ Chromatography__.

**Chromatography**is used to separate the**substances from a mixture.**

__Key Points__

**Chromatography**is a laboratory technique for the separation of a mixture.- It is used in analysis, isolation and purification, and it is commonly used in the chemical process industry as a component of small and large-scale production.
- The separation of a mixture by passing it in solution or suspension through a medium in which the components shift at different times is chromatography.
- There are
**several types of separation techniques**that are used in segregating a mixture of substances. Some of the common methods of separating substances:

**Handpicking****Sieving****Evaporation****Distillation****Threshing****Winnowing****Magnetic Separation****Filtration or Sedimentation****Separating Funnel**

Option 4 : 40 kJ/kg da

__Concept:__

From the conservation of energy, Energy of the mixture = Sum of the energy of the individual streams.

The energy carried by flowing steam = \(mass\;flow\;rate × \frac{{enthalpy}}{{unit\;mass}}\)

__Calculation:__

**Given:**

ṁ1 = 36 kg/min, ṁ2 = 14 kg/min, h_{1} = 36 kJ/(kg da), h_{2} = 50 kJ/(kg da)

Energy of the stream 1 = ṁ1 × h1 = 36 × 36 = 1296 kJ/(min da)

Energy of the stream 2 = ṁ2 × h2 = 14 × 50 = 700 kJ/(min da)

After the two streams mix into the one the resulting mass flow rate ṁ_{mix} = 36 + 14 = 50 kg/min

Now the energy of the resulting mass = h_{mix} × ṁf

From energy conservation E_{mix} = E1 + E2

⇒ hf × 50 = 1296 + 700

⇒ \({h_{mix}} = \frac{{1996}}{50} = 39.92 \;kJ/kg \;da\)

Option 2 : 2667 kJ/kg

__Concept:__

From the conservation of energy, Energy at input of mixing pipe = Energy at output of the mixing pipe

Energy carried by a flowing steam = \(mass\;flow\;rate \times \frac{{enthalpy}}{{unit\;mass}}\)

__Calculation:__

Energy of the stream 1 = ṁ_{1} × h_{1} = 1 × 3000 = 3000 kW

Energy of the stream 2 = ṁ_{2} × h_{2} = 2 × 2500 = 5000 kW

After the two streams mix into the one the resulting mass flow rate ṁ_{f }= 1 + 2 = 3 kg/s

Now the energy of the resulting mass = h_{f} × ṁ_{f}

From energy conservation E_{f }= E_{1} + E_{2}

⇒ h_{f} × 3 = 3000 + 5000

⇒ \({h_f} = \frac{{8000}}{3} = 2667 \;kJ/kg\)

A mixture of ideal gases has the following composition by mass:

N |
O |
CO |

60% |
30% |
10% |

If the Universal gas constant is 8314 J/kmol-K, the characteristic gas constant of the mixture (in J/kg.K) is _________.

**Concept:**

M_{T }R_{m} = m_{1}R_{1} + m_{2}R_{2 }+ m_{3}R_{3 }+ ..............+ m_{n}R_{n}

Rm = (m1R1 + m2R2 + m3R3 + ..............+ mnRn) / M_{T} ...................(1)

M_{T} = m1 + m2 + m_{3 }+...................+ m_{n}

R_{1, }R_{2}, R_{3 }are the characteristic gas constant of individual gases.

R_{m }= Characteristic gas constant of the mixture of gases

R (Characteristic gas constant) = (universal gas constant) / (molecular mass of gas)

R = R_{u }/ M

**Calculation:**

Given:

1,2, and 3 represent the gases N2, O2, and CO2 respectively.

m_{1 }= 0.6M_{T}, m_{2 }= 0.3M_{T},m_{3 }= 0.1M_{T}

R_{1 }= R_{u }/ 28, R_{2} = Ru / 32, R_{3 }= Ru / 44

R_{u} = 8314 J/kmol-k

Using equation 1,

\({R_m} = ({m_1}{R_1} + {m_2}{R_2} + {m_3}{R_3})/{M_T}\;\)

\({R_m} = \frac{{\left( {\frac{{0.6{M_T}{R_u}}}{{28}} + \frac{{0.3{M_T}{R_u}}}{{32}} + \frac{{0.1{M_T}{R_u}}}{{44}}} \right)}}{{{M_T}}}\)

\({R_m} = \frac{{0.6{R_u}}}{{28}} + \frac{{0.3{R_u}}}{{32}} + \frac{{0.1{R_u}}}{{44}}\)

R_{m} = 274.9961 J/kg-K

Option 2 : 1.21

__Explanation:__

For 1 mole of gas, we require 0.7 moles of air

\({\left( {{C_p}} \right)_{air}} = \left( {\frac{\gamma }{{\gamma - 1}}} \right)\bar R = 3.5\;\bar R\)

\({\left( {{C_v}} \right)_{air}} = \frac{{\bar R}}{{\left( {\gamma - 1} \right)}} = 2.5\;\bar R\)

**For freon,**

\({\left( {{C_p}} \right)_{freon}} = \left( {\frac{\gamma }{{\gamma - 1}}} \right)\bar R = 11\;\bar R\)

\({\left( {{C_v}} \right)_{freon}} = \frac{{\bar R}}{{\left( {\gamma - r} \right)}} = 10\;\bar R\)

\({\left( {{C_p}} \right)_{mix}} = \frac{{0.7 \times 3.5\;\bar R + 0.3 \times 11\;\bar R}}{1} = 5.75\;\bar R\)

\({\left( {{C_v}} \right)_{mix}} = \frac{{0.7 \times 2.5\;\bar R + 0.3 \times 10\;\bar R}}{1} = 4.75\;\bar R\)

∴ \({\gamma _{mix}} = \frac{{5.75\;\bar R}}{{4.75\;\bar R}}\)