Temperature of two moles of an ideal gas is increased by 300 K in a process, V = a T where, a is a positive constant. Find the work done by the gas in the given process.

We know that the work done by a gas in a thermodynamics process is, W = ∫ P d V . For an ideal gas, P V = n R T , so, W = ∫ n R T V d V . Given, n = 2 and V = a T , ∴ d V = - a T 2 d T . W = ∫ T 1 T 2 2 R T 2 a ( − a T 2 ) d T = - 2 R ∫ T 1 T 2 d T = - 2 R ( T 2 - T 1 ) . Given that, T 2 - T 1 = 300 K , ∴ W = - 600 R .

The pressure and the volume of a gas changes from, P 0 , V 0 to P 0 4 , 2 V 0 in a process, P V 2 = constant. Find the work done by the gas in the given process.

The thermodynamic process equation is given by, P V 2 = constant = k . We have to calculate the work done by the gas in the above thermodynamics process from, ( P 0 , V 0 ) to P 0 4 , 2 V 0 . The general formula of work done in thermodynamics process, W = ∫ P d V W = ∫ V 1 V 2 k V 2 d V (given that, P V 2 = k ) W = k ∫ V 1 V 2 d V V 2 = k − 1 V V 1 V 2 = k V 1 − k V 2 . We know that, P V 2 = P 1 V 1 2 = P 2 V 2 2 = k . Therefore, W = P 1 V 1 2 V 1 − P 2 V 2 2 V 2 W = P 1 V 1 - P 2 V 2 . Put the given values of ( P 1 , V 1 ) and ( P 2 , V 2 ) , W = P 0 V 0 - P 0 4 ( 2 V 0 ) W = P 0 V 0 2 .

E M B I B E

Understanding Physics JEE Main & Advanced WAVES AND THERMODYNAMICS>Chapter 5 - Laws of Thermodynamics>INTRODUCTORY EXERCISE 21.2

D. C. Pandey Solutions for Chapter: Laws of Thermodynamics, Exercise 2: INTRODUCTORY EXERCISE 21.2

Author:D. C. Pandey

D. C. Pandey Physics Solutions for Exercise - D. C. Pandey Solutions for Chapter: Laws of Thermodynamics, Exercise 2: INTRODUCTORY EXERCISE 21.2

Attempt the practice questions on Chapter 5: Laws of Thermodynamics, Exercise 2: INTRODUCTORY EXERCISE 21.2 with hints and solutions to strengthen your understanding. Understanding Physics JEE Main & Advanced WAVES AND THERMODYNAMICS solutions are prepared by Experienced Embibe Experts.

Questions from D. C. Pandey Solutions for Chapter: Laws of Thermodynamics, Exercise 2: INTRODUCTORY EXERCISE 21.2 with Hints & Solutions

EASY

JEE Advanced

IMPORTANT

Understanding Physics JEE Main & Advanced WAVES AND THERMODYNAMICS>Chapter 5 - Laws of Thermodynamics>INTRODUCTORY EXERCISE 21.2>Q 1

How many moles of Helium at temperature $300 K$ and pressure $1.00$ $atm$ are needed to make the internal energy of the gas $100 J$ ?

MEDIUM

JEE Advanced

IMPORTANT

Understanding Physics JEE Main & Advanced WAVES AND THERMODYNAMICS>Chapter 5 - Laws of Thermodynamics>INTRODUCTORY EXERCISE 21.2>Q 2

Temperature of four moles of a monoatomic gas is increased by $300 K$ in isochoric process. Find $W, Q$ and $Δ U$ .

MEDIUM

JEE Advanced

IMPORTANT

Understanding Physics JEE Main & Advanced WAVES AND THERMODYNAMICS>Chapter 5 - Laws of Thermodynamics>INTRODUCTORY EXERCISE 21.2>Q 3

Find the work done by the gas in the process $A B$ shown in the following figures:

Question Image

HARD

JEE Advanced

IMPORTANT

Understanding Physics JEE Main & Advanced WAVES AND THERMODYNAMICS>Chapter 5 - Laws of Thermodynamics>INTRODUCTORY EXERCISE 21.2>Q 4

Temperature of two moles of an ideal gas is increased by $300 K$ in a process, $V = \frac{a}{T}$ where, $a$ is a positive constant. Find the work done by the gas in the given process.

HARD

JEE Advanced

IMPORTANT

Understanding Physics JEE Main & Advanced WAVES AND THERMODYNAMICS>Chapter 5 - Laws of Thermodynamics>INTRODUCTORY EXERCISE 21.2>Q 5

The pressure and the volume of a gas changes from, $(P_{0}, V_{0})$ to $(\frac{P_{0}}{4}, 2 V_{0})$ in a process, $P V^{2} =$ constant. Find the work done by the gas in the given process.