Adiabatic Process

  • Calculate total work done in an adiabatic process.

Adiabatic Process Lines

A process in which the system is enclosed by adiabatic wall.

In an adiabatic process no heat is added to or removed from the gas (i.e., Q= 0). Examples include systems insulated so no heat is exchanged with the surroundings, and systems in which processes happen so fast that there is no time to add or remove heat. Because Q = 0 for an adiabatic process the First Law of Thermodynamics tells us that \(\Delta E = -W\). The energy for any work done comes from the change in the system’s internal energy. \[pv^\gamma = \mathrm{constant}\] where \(\gamma = \dfrac{C_p}{C_v}\) is the ratio of the heat capacity at constant pressure to the heat capacity at constant volume.

Homework for the topic Adiabatic Process

Solutions for the topic Adiabatic Process

Solved Example:

72-5-01

A process, in which the working substance neither receives nor gives out heat to its surroundings during its expansion or compression, is called: (VIZAG MT Mech 2015)

Correct Answer: C

Solved Example:

72-5-02

Adiabatic process is:

Correct Answer: B

Solved Example:

72-5-03

Work done in an adiabatic process between a given pair of end states depends on: (CIL MT Mech 2020)

Correct Answer: A

Solved Example:

72-5-04

An isolated system is one in which:

Correct Answer: B

Solved Example:

72-5-06

The ratio of specific heat at constant pressure (C$_p$) and specific heat at constant volume (C$_v$) is:

Correct Answer: C

Solved Example:

72-5-07

Characteristic gas constant of a gas is equal to: (ISRO Scientist ME 2010)

Solution:
Characteristics Gas Constant, \[R = C_p - C_v\]

Correct Answer: C

Solved Example:

72-5-08

The gas constant (R) is equal to the _________ of two specific heats. (SSC JE ME Paper 3- Sep 2019 Evening)

Correct Answer: D

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72-5-09

The value of gas constant (R) in S. I. units is: (RRC Group D Oct 2018 Shift 3)

Correct Answer: D

Solved Example:

72-5-10

When a perfect gas is heated at constant pressure from 15 $^\circ$C to 95$^\circ$ C, the heat required is 1136kJ/kg. When the same gas is heated at constant volume between the same temperatures the heat required is 808 kJ/kg. Calculate $\gamma$ for the gas.

Solution:
Constant Volume Process \[Q = 808\ kJ/kg\] \[T_1 = 15 + 273 = 288\ K,\] \[T_2 = 95 + 293 = 368\ K,\] \[\Delta T = T_2 - T_1 = 80\ K \] \[Q = mC_v \Delta T, 806 = 1 \times C_v \times 80, C_v = 10.1\ KJ/kg K\] Constant Pressure Process \begin{align*} Q &= 1136\ KJ/kg\\ Q &= m C_p \Delta T\\ 1136 &= 1 \times C_p \times 80\\ C_p &= 14.2\ kJ/kg K \end{align*} \[\gamma = \dfrac{C_p}{C_v} = \dfrac{14.2}{10.1} = 1.405\]

Correct Answer: B

Solved Example:

72-5-11

A mono-atomic ideal gas ($\gamma$ = 1.67, molecular weight = 40 ) is compressed adiabatically from 0.1 MPa , 300 K to 0.2 MPa . The universal gas constant is 8.314 kJ/KmolK. The work of compression of the gas is: (GATE ME 2010)

Solution:
Characteristic Gas Constant, \[ r = \dfrac{R}{M} = \dfrac{8.314}{40} = 0.208 kJ/kgK\] Using, \[\left(\dfrac{T_2}{T_1}\right) = \left(\dfrac{P_2}{P_1}\right)^{\dfrac{\gamma - 1}{\gamma}} \] \[ T_2 = 396.18K\] Work done in compression, \[= \dfrac{r(T_2 - T_1)}{\gamma - 1} = \dfrac{0.208 (396.18 - 300)}{1.67 - 1} = 29.8\]

Correct Answer: A

Solved Example:

72-5-12

5m$^3$ of air at 2 bar and 27$^\circ$C is compressed upto 6 bar. This process is governed by equation pV$^{1.3}$ = C. It is subsequently expanded to 2 bar. This process is reversible adiabatic. Find the net work in kJ:

Correct Answer: C