
Unit mass of a liquid with volume ${V_1}$ is completely changed into a gas of volume ${V_2}$ at a constant external pressure $P$ and temperature $T$ . If the latent heat of evaporation for the given mass is $L$ , then the increase in the internal energy of the system is?
A. Zero
B. $P({V_2} - {V_1})$
C. $L - P({V_2} - {V_1})$
D. $L$
Answer
135k+ views
Hint: Generally when external input energy is required to change the state from liquid into gases or vapor at constant temperature and constant external pressure is called the latent heat of vaporization. So, we can find out the increase internal energy of the system by using work done formula from the first law of thermodynamics, which is mentioned in below.
Formula used:
The formulae used in the solution of this problem are: -
$\Delta U = Q - W$
$Work Done = W = P\Delta V = P({V_2} - {V_1})$
Complete answer:
We know that the First Law of Thermodynamics is the application of conservation of energy and according to the first law: -
$\Delta U = Q - W$ … (1)
where $\Delta U = $ Change in Internal Energy
$Q = $ Heat Added to the system
$W = $ Work done by the system
Now, Pressure-Volume Work in thermodynamics is defined as: -
At constant pressure $Workdone = W = P\Delta V = P({V_2} - {V_1})$
From eq. (1), we get
$\Delta U = mL - P({V_2} - {V_1})$
$\left( {\therefore Q = m \times L} \right)$
where $L = $ Latent heat of evaporation (given)
and since the unit mass of liquid is taken, therefore $m = 1$
Thus, the increase in the internal energy of the system is: -
$ \Rightarrow \Delta U = (1)L - P({V_2} - {V_1}) = L - P({V_2} - {V_1})$
Hence, the correct option is (C) $L - P({V_2} - {V_1})$.
Note: Gases have higher internal energy as compared to solid and liquid state. This is because lot of energy is required to boind up the bonds between atoms or molecules and as a result this huge amount of energy provides a negative contribution to the internal energy in the solid and liquid state.
Formula used:
The formulae used in the solution of this problem are: -
$\Delta U = Q - W$
$Work Done = W = P\Delta V = P({V_2} - {V_1})$
Complete answer:
We know that the First Law of Thermodynamics is the application of conservation of energy and according to the first law: -
$\Delta U = Q - W$ … (1)
where $\Delta U = $ Change in Internal Energy
$Q = $ Heat Added to the system
$W = $ Work done by the system
Now, Pressure-Volume Work in thermodynamics is defined as: -
At constant pressure $Workdone = W = P\Delta V = P({V_2} - {V_1})$
From eq. (1), we get
$\Delta U = mL - P({V_2} - {V_1})$
$\left( {\therefore Q = m \times L} \right)$
where $L = $ Latent heat of evaporation (given)
and since the unit mass of liquid is taken, therefore $m = 1$
Thus, the increase in the internal energy of the system is: -
$ \Rightarrow \Delta U = (1)L - P({V_2} - {V_1}) = L - P({V_2} - {V_1})$
Hence, the correct option is (C) $L - P({V_2} - {V_1})$.
Note: Gases have higher internal energy as compared to solid and liquid state. This is because lot of energy is required to boind up the bonds between atoms or molecules and as a result this huge amount of energy provides a negative contribution to the internal energy in the solid and liquid state.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

What are examples of Chemical Properties class 10 chemistry JEE_Main

JEE Main 2025 Session 2 Schedule Released – Check Important Details Here!

JEE Main 2025 Session 2 Admit Card – Release Date & Direct Download Link

JEE Main 2025 Session 2 Registration (Closed) - Link, Last Date & Fees

JEE Mains Result 2025 NTA NIC – Check Your Score Now!

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Degree of Dissociation and Its Formula With Solved Example for JEE

A body is falling from a height h After it has fallen class 11 physics JEE_Main

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Elastic Collisions in One Dimension - JEE Important Topic

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion In A Plane: Line Class 11 Notes: CBSE Physics Chapter 3
