Heat, Internal Energy And Work - Thermodynamics

Quantities Involved in the First Law of Thermodynamics

Heat (Q): It is the energy that is transferred between a system and its environment because of the temperature gradient.

There are two types of internal energy

Types of internal energy

Due to molecular motion internal energy is kinetic internal energy (UK).

Due to molecular configuration, it is called internal potential energy (UP).

Important points :

1. Heat and work are path-dependent quantities and Internal energy is point function.

2. $\begin{aligned} & \Delta W=P \Delta V=P\left(V_f-V_i\right) \\ & \Delta W=\text { positive if } V_f>V_i \text { i.e. system expands against some external force. } \\ & \Delta W=\text { negative if } V_f<V_i \text { i.e. system contracts because of some external force exerted by the surrounding. }\end{aligned}$

3. The area of the P-v diagram on the volume axis gives the work done in a reversible process. Also for quasistatic process work is given by

$W=\int_{V_1}^{V_2} P . d V$

4. And for a cyclic process the clockwise area will show positive work and the anticlockwise area will show negative work done.

5. The internal energy of an ideal gas is totally kinetic and is given by

$U=\frac{3}{2} \cdot n \cdot R \cdot T$

So, the Internal energy of an ideal gas is the function of temperature only.

6. For heat transfer

$\Delta Q=m L$ (for change of state)
$\Delta Q=m s \Delta T$ (for change in temperature)
or,
$
\Delta Q=n c \Delta T
$

Where, c = molar specific heat capacity

Sign of dQ(Heat)

$d Q>0$ if heat is given to the system
$d Q<0$ if heat is extracted from the system

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Solved Examples Based on Heat, Internal energy and Work in Thermodynamics

Example 1: In a given process of an ideal gas dW=0 and dQ<0. Then for the gas

1) The temperature will decrease

2) The pressure will increase

3) The volume will increase

4) The temperature will increase

Solution:

From the question and by applying the first law of thermodynamics

$
\Delta Q=\Delta U+\Delta W
$

Now
$\Delta W=0, \quad$ So, $\quad \Delta Q=\Delta U$
And $\Delta Q<0$, So, $\Delta U<0$,
But for an ideal gas $U \alpha T \Rightarrow d T<0$

Hence temperature will decrease.

Hence, the answer is the option 1.

Example 2: As shown in the figure, an ideal gas is taken through the cycle ABCA. If the net heat supplied in the cycle is 3 J. The work done by the gas in the process A->B is (in J)

1) -1

2) 5

3) 2

4) 3

Solution:

Process B->C at constant volume $W_{B C}=0$
Process C->A at constant pressure $W_{C A}=P \Delta V=2(6-4)=4 J$
As the volume of gas increases work done is positive.
As the entire process is cyclic.
$
\begin{aligned}
& \text { so } \Delta U=0 \\
& \Delta Q=\Delta W+\Delta U \Rightarrow W_{A B}+W_{B C}+W_{C A} \\
& 3=W_{A B}+0+4 \\
& W_{A B}=-1 J
\end{aligned}
$

Example 3: When 30J of work was done on gas, 20J of heat energy was released. If the initial energy of the gas was 40J, what is the final internal energy (in J)?

1) 50

2) 30

3) 40

4) 20

Solution:

Sign of dU (internal energy)

If internal energy increases then dU is positive, if internal energy decreases then dU is negative.

$
\begin{aligned}
& \left.d U>0 \text { (if } U_f>U_i\right) \\
& d U<0 \text { (if } U_f<U_i \text { ) } \\
& \Delta U=Q-W=(-20)-(-30)=10 J
\end{aligned}
$
$\Delta U$ is positive, so internal energy is increasing.
$
\begin{aligned}
& \Delta U=U_f-U_i=10 \Rightarrow U_f-40=10 \\
& U_f=50 \mathrm{~J}
\end{aligned}
$

Hence, the answer is 50.

Example 4: The change in internal energy (in \%) when a gas is cooled from $927^{\circ}$ to $27^{\circ} \mathrm{C}$, is:

1) 75

2) 200

3) 250

4) 300

Solution:

T₁ = 927 + 273 = 1200 K

T₂ = 27 + 273 = 300 K

Internal energy ∝ Temperature

U ∝ T

So,

(U₁ / U₂) = (T₁ / T₂)

[(U₁ – U₂) / U₂] = [(T₁ – T₂) / T₂]

% change in energy = [(T₁ – T₂) / T₂] X 100

% change in energy = [{(1200 – 300) / (300)} × (100)] = 3 × 100 = 300%

Hence, the answer is 300.

Example 5: An equal amount of an ideal monoatomic gas at 400K is filled in two cylinders A and B. The piston A is free to move while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 45 K. Then the rise in temperature of gas in B is (In Kelvin)

1) 75

2) 100

3) 55

4) 45

Solution:

The gas filled in monoatomic piston A is free to move i.e. it is an isobaric process. Piston B is fixed i.e. it is an isochoric process. If the same amount of heat is given,

$\begin{aligned}(\Delta Q)_{\text {isobaric }} & =(\Delta Q)_{\text {isochoric }} \\ \mu C_p(\Delta T)_A & =\mu C_v(\Delta T)_B \\ (\Delta T)_B & =\left(\frac{C_P}{C_v}\right)(\Delta T)_A \\ & =\gamma \times(\Delta T)_A \\ & =\frac{5}{3} \times 45=75 K\end{aligned}$

Hence, the answer is 75.

Summary

In our daily lives, heat and energy are used in a variety of physical, chemical, and biological processes. Thus, it is imperative that you study them. Temperature and heat are frequently used interchangeably. This lesson, however, distinguishes clearly between the two and explains them. We also studied the causes and effects of an object's thermal expansion. In addition, we gained an understanding of the labour done and the internal energy of a system. You might study the connection between enthalpy and internal energy.