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Adiabatic Process - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Adiabatic process is considered one the most difficult concept.

  • 68 Questions around this concept.

Solve by difficulty

When a gas expands adiabatically 

A given system undergoes a change in which the work done by the system equals the decrease in its internal energy. The system must have undergone an 

During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio C_{P}/C_{V} for the gas is

In a P-V diagram for an ideal gas for different process is as shown . In graph curve OR represents 

For an adiabatic process which of the following statement is not true 

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For a thermodynamic process, specific heat of gas is zero. The process is 

If \Delta U and \Delta W represent the increase in internal energy and work done by the system respectively in a thermodynamical process, which of the following is true?

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In thermodynamic processes, which of the following statements is not true?

The internal energy change in a system that has absorbed 2 kcals of heat and done 500 J of work is;

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A monoatomic gas at a pressure P, having a volume V expands isothermally to a volume 2 V, and then adiabatically to a volume 16 V. The final pressure of the gas is: 

take\:\:\left( \gamma= \frac{5}{3} \right ) 

Concepts Covered - 1

Adiabatic process

Adiabatic process - 

When a thermodynamic system undergoes a process, such that there is no exchange of heat takes place between the system and surroundings, this process is known as the adiabatic process.

In this process P, V and T changes but  \Delta Q = 0.

From first law of thermodynamics - 

             \Delta Q=\Delta U+\Delta W

Now in adiabatic - 

                 0=\Delta U+\Delta W

      So, \Delta U=-\Delta W for adiabatic process

Now, let us take two cases, first is for expansion in which the work done is positive and second one is compression in which the work done is negative - 

                    \begin{array}{l}{\text { If } \Delta W=\text { positive then } \Delta U \ \textup{become}\ \text { negative so temperature decreases ie.,}} {\text { adiabatic expansion produce cooling. }} \\ {\text { If } \Delta W \text { = negative then } \Delta U\ \textup{become} \ \text { positive so temperature increases ie., }} {\text { adiabatic compression produce heating. }}\end{array}


Equations of Adiabatic process - 

1. P V^{\gamma}=\text { constant; where } \gamma=\frac{C_{P}}{C_{V}}    - - - - Relating Pressure and volume

2. T {V}^{\gamma - 1}=\text { constant } \Rightarrow T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma-1} \text { or } T \propto V^{1-\gamma}  - - - - -Relating Temperature and volume

3. \frac{T^{\gamma}}{P^{\gamma-1}}=\text { const. } \Rightarrow T_{1}^{\gamma} P_{1}^{1-\gamma}=T_{2}^{\gamma} P_{2}^{1-\gamma} \text { or } T \propto P^{\frac{\gamma-1}{\gamma}} \text { or } P \propto T^{\frac{7}{\gamma-1}} 


For the slope of adiabatic curve on PV curve, we have to differentiate the adiabatic relation - 

              As, P V^{\gamma}=\text { constant }


                  \begin{array}{l}{d P V^{\gamma}+P \gamma V^{\gamma-1} d V=0} \\ \\ {\frac{d P}{d V}=-\gamma \frac{P V^{\gamma-1}}{V^{\gamma}}=-\gamma\left(\frac{P}{V}\right)}\end{array}

So we can say that in the given graph, the slope  =   \tan \phi=-\gamma\left(\frac{P}{V}\right)            

Also, we have studied that the slope of the isothermal curve on PV diagram is  =   \frac{-P}{V}

So, we can say that the - \text { (Slope) }_{a d iabatic}=\gamma \times(\text { Slope })_{isothermal}\ , \text {or } \frac{(\text { Slope })_{a d iabatic}}{(\text { Slope })_{isothermal}}>1

With the help of graph we can see that the adiabatic curve is more steeper than the isothermal curve- 


Specific heat in adiabatic process - Specific heat of a gas during adiabatic change is zero. Mathematically  -

                                                             C=\frac{Q}{m \Delta T}=\frac{0}{m \Delta T}=0 \quad[\text { As } Q=0]

Note- Even though heat is not supplied or taken out during the process but still, the temperature change is taking place. So we can say that Specific heat for an adiabatic process is zero.

Work done in adiabatic process - 

                                    W=\int_{V_{i}}^{V_{f}} P d V=\int_{V_{i}}^{V_{f}} \frac{K}{V^{\gamma}} d V \Rightarrow W=\frac{\left[P_{i} V_{i}-P_{f} V_{f}\right]}{(\gamma-1)}=\frac{\mu R\left(T_{i}-T_{f}\right)}{(\gamma-1)}

So, if \gamma is increasing then work done will be decreasing. As we know that - 

                                  \because \gamma_{\text {mono}}>\gamma_{d iatomic}>\gamma_{triatomic } \Rightarrow W_{mono}<W_{d iatomic} <W_{triatomic}


Elasticity in adiabatic process - As,  P V^{\gamma}=\text { constant }


                                                         \begin{array}{l}{\text { Differentiating both sides of the above equation, we get - } \ d P V^{\gamma}+P \gamma V^{\gamma-1} d V=0} \\ \\ {\qquad \gamma P=\frac{d P}{-d V / V}=\frac{\text { Stress }}{\text { Strain }}=E_{\phi} \Rightarrow E_{\phi}=\gamma P} \\ \\ {\text { i.e. adiabatic elasticity is } \gamma \text { times that of pressure }} \\ \\ {\qquad E_{\theta}=P \Rightarrow \frac{E_{\phi}}{E_{\theta}}=\gamma=\frac{C_{P}}{C_{V}}} \\ \\ {\text { i.e. the ratio of two elasticity of gases is equal to the ratio of two }} \\ {\text { specific heats. }}\end{array}


                                                          E_{\phi}= \text {Elasticity in adiabatic process} \\ and \\ E_{ \theta }= \text {Elasticity in isothermal process}

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Reference Books

Adiabatic process

Physics Part II Textbook for Class XI

Page No. : 312

Line : 15

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