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Thermodynamics

148230 A gaseous mixture consists of $16 g$ of helium and $16 g$ of oxygen. The ratio $C_{p} / C_{V}$ of the mixture is

1 1.4

2 1.54

3 1.59

4 1.62

Explanation:

D For $16 g$ of helium, $n_{1} = \frac{16}{4} = 4$
For $16 g$ of oxygen $n_{2} = \frac{16}{32} = \frac{1}{2}$
For mixture of gases,
$C_{V} = \frac{n_{1} C_{V_{1}} + n_{2} C_{v_{2}}}{n_{1} + n_{2}}$
Where, $C_{V} = \frac{f}{2} R$
For Helium, $f = 3$ then $C_{V_{1}} = \frac{3}{2} R$
For oxygen, $f = 5$ then $C_{V_{2}} = \frac{5}{2} R$
$C_{V} = \frac{4 \times \frac{3}{2} R + \frac{1}{2} \times \frac{5}{2} R}{4 + \frac{1}{2}}$
$= \frac{6 R + \frac{5}{4} R}{\frac{9}{2}}$
$= \frac{29 R \times 2}{9 \times 4} = \frac{29 R}{18}$
We know,
$C_{v} = \frac{R}{γ - 1}$
$γ - 1 = \frac{R}{C_{v}}$
$γ = \frac{C_{p}}{C_{v}} = \frac{R}{C_{v}} + 1$
$= \frac{R}{\frac{29}{18}} R$
$= 1.62$

COMEDK 2020

Thermodynamics

148231 In a thermodynamic system, $Q$ represents the energy transferred to or from a system by heat and $W$ represents the energy transferred to or from a system by work
I. $Q > 0$ and $W = 0$
II. $Q < 0$ and $W = 0$
III. $W > 0$ and $Q = 0$
IV. $W < 0$ and $Q = 0$
Which of the above will lead to an increase in the internal energy of the system?

1 I only

2 II only

3 I and IV only

4 II and III only

5 II and IV only

Explanation:

C According first law of thermodynamics,
$Q = Δ U + W$
$Δ U = Q - W$
According to given conditions
I. $Q > 0$ and $W = 0$
$or Δ U = Q$
Here $Q > 0$ , therefore $Δ U$ will be increase.
II. $Q < 0$ and $W = 0$
$or Δ U = Q$
Here, $Q < 0$ , therefore $Δ U$ will be decrease.
III. $W > 0$ and $Q = 0$
$or Δ U = - W$
Here, $W > 0$ therefore $Δ U$ will be decrease.
IV. $W < 0$ and $Q = 0$
$or Δ U = 0 - (- W)$
$Δ U = W$
Here, $W < 0$ , therefore $Δ U$ will be increase. Hence, condition I and IV will lead to an increase in the internal energy of the system. Therefore, option (c) is correct.

Kerala CEE -2018

Thermodynamics

148233 The P-V diagram of a gas system undergoing cyclic process is shown here. The work done during isobaric compression is

1

100 J

2

200 J

3

600 J

4

500 J

5

400 J

Explanation:

E
For isobaric compression, $(D \to A)$
Work done $= - PdV$ (Anti clock wise)
$= - 2 \times 10^{2} \times (1 - 3)$
$= (- 2) \times 10^{2} \times (- 2)$
$= 4 \times 10^{2} = 400 J$

Kerala CEE- 2013

Thermodynamics

148234 Identify the wrong statement

1 For isothermal process,

Δ T = 0

2 For isochoric process,

Δ V = 0

3 For isobaric process,

Δ P = 0

4 For adiabatic process,

Δ Q = 0

5 For cyclic process,

Δ W = 0

Explanation:

E
For cyclic process,
After undergoing multiple processes in one cycle system returns back to its initial state.
$\sum_{i = 1}^{n} Q_{i} = \sum_{i = 1}^{n} W_{i}$
$Q_{n e t} = W_{n e t}$

Kerala CEE 2012

Thermodynamics

148230 A gaseous mixture consists of $16 g$ of helium and $16 g$ of oxygen. The ratio $C_{p} / C_{V}$ of the mixture is

1 1.4

2 1.54

3 1.59

4 1.62

Explanation:

D For $16 g$ of helium, $n_{1} = \frac{16}{4} = 4$
For $16 g$ of oxygen $n_{2} = \frac{16}{32} = \frac{1}{2}$
For mixture of gases,
$C_{V} = \frac{n_{1} C_{V_{1}} + n_{2} C_{v_{2}}}{n_{1} + n_{2}}$
Where, $C_{V} = \frac{f}{2} R$
For Helium, $f = 3$ then $C_{V_{1}} = \frac{3}{2} R$
For oxygen, $f = 5$ then $C_{V_{2}} = \frac{5}{2} R$
$C_{V} = \frac{4 \times \frac{3}{2} R + \frac{1}{2} \times \frac{5}{2} R}{4 + \frac{1}{2}}$
$= \frac{6 R + \frac{5}{4} R}{\frac{9}{2}}$
$= \frac{29 R \times 2}{9 \times 4} = \frac{29 R}{18}$
We know,
$C_{v} = \frac{R}{γ - 1}$
$γ - 1 = \frac{R}{C_{v}}$
$γ = \frac{C_{p}}{C_{v}} = \frac{R}{C_{v}} + 1$
$= \frac{R}{\frac{29}{18}} R$
$= 1.62$

COMEDK 2020

Thermodynamics

148231 In a thermodynamic system, $Q$ represents the energy transferred to or from a system by heat and $W$ represents the energy transferred to or from a system by work
I. $Q > 0$ and $W = 0$
II. $Q < 0$ and $W = 0$
III. $W > 0$ and $Q = 0$
IV. $W < 0$ and $Q = 0$
Which of the above will lead to an increase in the internal energy of the system?

1 I only

2 II only

3 I and IV only

4 II and III only

5 II and IV only

Explanation:

C According first law of thermodynamics,
$Q = Δ U + W$
$Δ U = Q - W$
According to given conditions
I. $Q > 0$ and $W = 0$
$or Δ U = Q$
Here $Q > 0$ , therefore $Δ U$ will be increase.
II. $Q < 0$ and $W = 0$
$or Δ U = Q$
Here, $Q < 0$ , therefore $Δ U$ will be decrease.
III. $W > 0$ and $Q = 0$
$or Δ U = - W$
Here, $W > 0$ therefore $Δ U$ will be decrease.
IV. $W < 0$ and $Q = 0$
$or Δ U = 0 - (- W)$
$Δ U = W$
Here, $W < 0$ , therefore $Δ U$ will be increase. Hence, condition I and IV will lead to an increase in the internal energy of the system. Therefore, option (c) is correct.

Kerala CEE -2018

Thermodynamics

148233 The P-V diagram of a gas system undergoing cyclic process is shown here. The work done during isobaric compression is

1

100 J

2

200 J

3

600 J

4

500 J

5

400 J

Explanation:

E
For isobaric compression, $(D \to A)$
Work done $= - PdV$ (Anti clock wise)
$= - 2 \times 10^{2} \times (1 - 3)$
$= (- 2) \times 10^{2} \times (- 2)$
$= 4 \times 10^{2} = 400 J$

Kerala CEE- 2013

Thermodynamics

148234 Identify the wrong statement

1 For isothermal process,

Δ T = 0

2 For isochoric process,

Δ V = 0

3 For isobaric process,

Δ P = 0

4 For adiabatic process,

Δ Q = 0

5 For cyclic process,

Δ W = 0

Explanation:

E
For cyclic process,
After undergoing multiple processes in one cycle system returns back to its initial state.
$\sum_{i = 1}^{n} Q_{i} = \sum_{i = 1}^{n} W_{i}$
$Q_{n e t} = W_{n e t}$

Kerala CEE 2012

Thermodynamics

148230 A gaseous mixture consists of $16 g$ of helium and $16 g$ of oxygen. The ratio $C_{p} / C_{V}$ of the mixture is

1 1.4

2 1.54

3 1.59

4 1.62

Explanation:

D For $16 g$ of helium, $n_{1} = \frac{16}{4} = 4$
For $16 g$ of oxygen $n_{2} = \frac{16}{32} = \frac{1}{2}$
For mixture of gases,
$C_{V} = \frac{n_{1} C_{V_{1}} + n_{2} C_{v_{2}}}{n_{1} + n_{2}}$
Where, $C_{V} = \frac{f}{2} R$
For Helium, $f = 3$ then $C_{V_{1}} = \frac{3}{2} R$
For oxygen, $f = 5$ then $C_{V_{2}} = \frac{5}{2} R$
$C_{V} = \frac{4 \times \frac{3}{2} R + \frac{1}{2} \times \frac{5}{2} R}{4 + \frac{1}{2}}$
$= \frac{6 R + \frac{5}{4} R}{\frac{9}{2}}$
$= \frac{29 R \times 2}{9 \times 4} = \frac{29 R}{18}$
We know,
$C_{v} = \frac{R}{γ - 1}$
$γ - 1 = \frac{R}{C_{v}}$
$γ = \frac{C_{p}}{C_{v}} = \frac{R}{C_{v}} + 1$
$= \frac{R}{\frac{29}{18}} R$
$= 1.62$

COMEDK 2020

Thermodynamics

148231 In a thermodynamic system, $Q$ represents the energy transferred to or from a system by heat and $W$ represents the energy transferred to or from a system by work
I. $Q > 0$ and $W = 0$
II. $Q < 0$ and $W = 0$
III. $W > 0$ and $Q = 0$
IV. $W < 0$ and $Q = 0$
Which of the above will lead to an increase in the internal energy of the system?

1 I only

2 II only

3 I and IV only

4 II and III only

5 II and IV only

Explanation:

C According first law of thermodynamics,
$Q = Δ U + W$
$Δ U = Q - W$
According to given conditions
I. $Q > 0$ and $W = 0$
$or Δ U = Q$
Here $Q > 0$ , therefore $Δ U$ will be increase.
II. $Q < 0$ and $W = 0$
$or Δ U = Q$
Here, $Q < 0$ , therefore $Δ U$ will be decrease.
III. $W > 0$ and $Q = 0$
$or Δ U = - W$
Here, $W > 0$ therefore $Δ U$ will be decrease.
IV. $W < 0$ and $Q = 0$
$or Δ U = 0 - (- W)$
$Δ U = W$
Here, $W < 0$ , therefore $Δ U$ will be increase. Hence, condition I and IV will lead to an increase in the internal energy of the system. Therefore, option (c) is correct.

Kerala CEE -2018

Thermodynamics

148233 The P-V diagram of a gas system undergoing cyclic process is shown here. The work done during isobaric compression is

1

100 J

2

200 J

3

600 J

4

500 J

5

400 J

Explanation:

E
For isobaric compression, $(D \to A)$
Work done $= - PdV$ (Anti clock wise)
$= - 2 \times 10^{2} \times (1 - 3)$
$= (- 2) \times 10^{2} \times (- 2)$
$= 4 \times 10^{2} = 400 J$

Kerala CEE- 2013

Thermodynamics

148234 Identify the wrong statement

1 For isothermal process,

Δ T = 0

2 For isochoric process,

Δ V = 0

3 For isobaric process,

Δ P = 0

4 For adiabatic process,

Δ Q = 0

5 For cyclic process,

Δ W = 0

Explanation:

E
For cyclic process,
After undergoing multiple processes in one cycle system returns back to its initial state.
$\sum_{i = 1}^{n} Q_{i} = \sum_{i = 1}^{n} W_{i}$
$Q_{n e t} = W_{n e t}$

Kerala CEE 2012

Thermodynamics

148230 A gaseous mixture consists of $16 g$ of helium and $16 g$ of oxygen. The ratio $C_{p} / C_{V}$ of the mixture is

1 1.4

2 1.54

3 1.59

4 1.62

Explanation:

D For $16 g$ of helium, $n_{1} = \frac{16}{4} = 4$
For $16 g$ of oxygen $n_{2} = \frac{16}{32} = \frac{1}{2}$
For mixture of gases,
$C_{V} = \frac{n_{1} C_{V_{1}} + n_{2} C_{v_{2}}}{n_{1} + n_{2}}$
Where, $C_{V} = \frac{f}{2} R$
For Helium, $f = 3$ then $C_{V_{1}} = \frac{3}{2} R$
For oxygen, $f = 5$ then $C_{V_{2}} = \frac{5}{2} R$
$C_{V} = \frac{4 \times \frac{3}{2} R + \frac{1}{2} \times \frac{5}{2} R}{4 + \frac{1}{2}}$
$= \frac{6 R + \frac{5}{4} R}{\frac{9}{2}}$
$= \frac{29 R \times 2}{9 \times 4} = \frac{29 R}{18}$
We know,
$C_{v} = \frac{R}{γ - 1}$
$γ - 1 = \frac{R}{C_{v}}$
$γ = \frac{C_{p}}{C_{v}} = \frac{R}{C_{v}} + 1$
$= \frac{R}{\frac{29}{18}} R$
$= 1.62$

COMEDK 2020

Thermodynamics

148231 In a thermodynamic system, $Q$ represents the energy transferred to or from a system by heat and $W$ represents the energy transferred to or from a system by work
I. $Q > 0$ and $W = 0$
II. $Q < 0$ and $W = 0$
III. $W > 0$ and $Q = 0$
IV. $W < 0$ and $Q = 0$
Which of the above will lead to an increase in the internal energy of the system?

1 I only

2 II only

3 I and IV only

4 II and III only

5 II and IV only

Explanation:

C According first law of thermodynamics,
$Q = Δ U + W$
$Δ U = Q - W$
According to given conditions
I. $Q > 0$ and $W = 0$
$or Δ U = Q$
Here $Q > 0$ , therefore $Δ U$ will be increase.
II. $Q < 0$ and $W = 0$
$or Δ U = Q$
Here, $Q < 0$ , therefore $Δ U$ will be decrease.
III. $W > 0$ and $Q = 0$
$or Δ U = - W$
Here, $W > 0$ therefore $Δ U$ will be decrease.
IV. $W < 0$ and $Q = 0$
$or Δ U = 0 - (- W)$
$Δ U = W$
Here, $W < 0$ , therefore $Δ U$ will be increase. Hence, condition I and IV will lead to an increase in the internal energy of the system. Therefore, option (c) is correct.

Kerala CEE -2018

Thermodynamics

148233 The P-V diagram of a gas system undergoing cyclic process is shown here. The work done during isobaric compression is

1

100 J

2

200 J

3

600 J

4

500 J

5

400 J

Explanation:

E
For isobaric compression, $(D \to A)$
Work done $= - PdV$ (Anti clock wise)
$= - 2 \times 10^{2} \times (1 - 3)$
$= (- 2) \times 10^{2} \times (- 2)$
$= 4 \times 10^{2} = 400 J$

Kerala CEE- 2013

Thermodynamics

148234 Identify the wrong statement

1 For isothermal process,

Δ T = 0

2 For isochoric process,

Δ V = 0

3 For isobaric process,

Δ P = 0

4 For adiabatic process,

Δ Q = 0

5 For cyclic process,

Δ W = 0

Explanation:

E
For cyclic process,
After undergoing multiple processes in one cycle system returns back to its initial state.
$\sum_{i = 1}^{n} Q_{i} = \sum_{i = 1}^{n} W_{i}$
$Q_{n e t} = W_{n e t}$

Kerala CEE 2012

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