- For an adiabatic process of ideal gas equation we have

$PV^{\gamma} = K$ Where $\gamma$ is the ratio of specific heat (ordinary or molar) at constant pressure and at constant volume

$\gamma = \frac {C_p}{C_v}$

- Suppose in an adiabatic process pressure and volume of a sample of gas changes from (P
_{1}, V_{1}) to (P_{2}, V_{2}) then we have

Thus, $P = \frac {K}{V^{\gamma}}$

- Work done by gas in this process is

$W = \int_{V_1}^{V_2} PdV$

where limits of integration goes from V_{1}to V_{2}

Putting for $P = \frac {K}{V^{\gamma}}$, and integrating we get,

$W=\int_{V_1}^{V_2} \frac {K}{V^{\gamma}} dV$

$W= \frac {C}{1 - \gamma} (\frac {1}{V_2^{\gamma -1}} - \frac {1}{V_1^{\gamma -1}})$

Now we have $P_1V_1^{\gamma}=P_2V_2^{\gamma}=K$

$W= \frac {1}{1 - \gamma} (P_2V_2 -P_1V_1)$

- Now Since $P_1V_1=nRT_1$ and $P_2V_2=nRT_2$ , Work done can be expressed as

$W= \frac {nR}{1 - \gamma} (T_1 - T_2)$

- In and adiabatic process if W>0 i.e., work is done by the gas then T
_{2}< T_{1}

- If work is done on the gas (W<0) then T
_{2}> T_{1}i.e., temperature of gas rises.

Given $4^{1.4} =6.96$

For an adiabatic Process

$PV^{\gamma} = K$

Here $P_1=4 \times 10^5 N/m^2$ , $V_1=2m^3$ ,$V_2=.5 m^3$, $P_2=?$

Now $P_1V_1^{\gamma} = P_2V_2^{\gamma}$

or

$P_2 = P_1 (\frac {V_1}{V_2})^{\gamma}= (4 \times 10^5) (\frac {2}{.5})^{\gamma}$

$=2.78 \times 10^6 N/m^2$

Now Work done in an adiabatic process

$W= \frac {1}{1 - \gamma} (P_2V_2 -P_1V_1)$

Substituting all the values , we have

$W=-14.75 \times 10^5$ J

Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in below figure.

(a) Find the work done when the gas is taken from state 1 to state 2.

(b) What is the ratio of temperature $ \frac {T_1}{T_2}$, if $V_2 = 2V_1$?

(c) Given the internal energy for one mole of gas at temperature T is $ \frac {3}{2} RT$, find the heat supplied to the gas when it is taken from state 1 to 2, with $V_2 = 2V_1$

This is clearly an adiabatic Process with $\gamma =\frac {1}{2}$

$PV^{1/2} = Constant$

a. Now work done is given by

$W= \frac {1}{1 - \gamma} (P_2V_2 -P_1V_1)$

or

$W= 2(P_2V_2 -P_1V_1)$

b. Now since $PV=RT$ or $P =\frac {RT}{V}$

So, $PV^{1/2} = Constant$ becomes

$\frac {RT}{V} V^{1/2} = Constant$

$TV^{-1/2} = constant$

Therefore

$T_1 V_1^{-1/2}= T_2 V_2^{-1/2}$

$\frac {T_1}{T_2} = (\frac {V_2}{V_1})^{-1/2}$

or

$\frac {T_1}{T_2}= \frac {1}{\sqrt {2}}$

c. Change in Internal Energy will be given by

$\Delta U = U_2 -U_1 = \frac {3}{2} RT_2 - \frac {3}{2} RT_1= \frac {3}{2}R(T_2 -T_1) = \frac {3}{2}R T_1( \sqrt {2} -1)$

$\Delta W= \frac {nR}{1 - \gamma} (T_1 - T_2) =2R T_1(\sqrt {2} -1)$

Now

$\Delta Q = \Delta U + \Delta W= \frac {3}{2}R T_1( \sqrt {2} -1) + 2R T_1(\sqrt {2} -1)=\frac {7}{2} RT_1(\sqrt {2} -1)$

what all is true for a adiabatic process

a. $dQ=0$

b. $dW=-nC_vdT$

c. $TV^{\gamma -1}=constant$

d. all the above

(d). As all of them are true

- Introduction
- Concept of Heat
- P-V Indicator Digram
- |
- Work done by Gas in volume changes
- |
- Internal Energy
- |
- First Law of Thermodynamics
- |
- Specific Heat Capacity of Ideal GAS
- |
- Thermodynamic Processes
- |
- Quasi static Processes
- |
- Isothermal Process
- |
- Adiabatic Process
- |
- Isochoric process
- |
- Isobaric process
- |
- Cyclic process
- |
- Work done in Isothermal process
- |
- Work done in an Adiabatic process
- |
- Heat Engine and efficiency
- |
- Principle of a Refrigerator
- |
- Second law of thermodynamics
- |
- Reversibility and irreversibility
- |
- Carnot's Heat Engine
- |
- Carnot Theorem
- |
- Solved Examples

- Thermodynamics Questions
- |
- Multiple Choice Questions
- |
- P-V diagram Problems and Solutions
- |
- Carnot Cycle Problems

Class 11 Maths Class 11 Physics Class 11 Chemistry

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