If two bodies are not in thermal equilibrium, then in which direction the heat will flow?

If two bodies are not in thermal equilibrium, then the heat flows from body at high temperature to the body at low temperature. 

The flow of heat takes place for the bodies which are not in thermal equilibrium. 
449 Views

A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500 K It is desired to have an engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be

  • the efficiency of Carnot engine cannot be made larger than 50%

  • 1200 K

  • 750 K 

  • 750 K 


C.

750 K 

Efficiency
straight eta space equals space 1 minus straight T subscript sink over straight T subscript source
Now comma space 0.4 space equals 1 minus fraction numerator straight T subscript sink over denominator 500 straight K end fraction
rightwards double arrow space straight T subscript sink space equals space 0.6 space straight x space 500 space straight K
space equals space 300 space straight K
Thus comma space 0.6 space equals space 1 minus fraction numerator 300 space straight K over denominator straight T apostrophe subscript source end fraction
rightwards double arrow space straight T to the power of apostrophe subscript source space equals space fraction numerator 300 space straight K over denominator 0.4 end fraction
space equals 750 space straight K

Efficiency
straight eta space equals space 1 minus straight T subscript sink over straight T subscript source
Now comma space 0.4 space equals 1 minus fraction numerator straight T subscript sink over denominator 500 straight K end fraction
rightwards double arrow space straight T subscript sink space equals space 0.6 space straight x space 500 space straight K
space equals space 300 space straight K
Thus comma space 0.6 space equals space 1 minus fraction numerator 300 space straight K over denominator straight T apostrophe subscript source end fraction
rightwards double arrow space straight T to the power of apostrophe subscript source space equals space fraction numerator 300 space straight K over denominator 0.4 end fraction
space equals 750 space straight K

690 Views

The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat extracted from the source in a single cycle is

  • povo

  • open parentheses 13 over 2 close parentheses straight p subscript 0 straight v subscript 0
  • open parentheses 11 over 2 close parentheses straight p subscript straight o straight v subscript 0
  • open parentheses 11 over 2 close parentheses straight p subscript straight o straight v subscript 0

B.

open parentheses 13 over 2 close parentheses straight p subscript 0 straight v subscript 0

Heat is extracted from the source in path DA and AB is

space straight Q space equals space 3 over 2 straight R open parentheses fraction numerator straight P subscript 0 straight V subscript 0 over denominator straight R end fraction close parentheses space plus 5 over 2 straight R open parentheses fraction numerator begin display style 2 straight P subscript 0 straight V subscript 0 end style over denominator straight R end fraction close parentheses space equals space 13 over 2 straight P subscript 0 straight V subscript 0

Heat is extracted from the source in path DA and AB is

space straight Q space equals space 3 over 2 straight R open parentheses fraction numerator straight P subscript 0 straight V subscript 0 over denominator straight R end fraction close parentheses space plus 5 over 2 straight R open parentheses fraction numerator begin display style 2 straight P subscript 0 straight V subscript 0 end style over denominator straight R end fraction close parentheses space equals space 13 over 2 straight P subscript 0 straight V subscript 0

565 Views

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is V0 and its pressure is P0. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency

  • fraction numerator 1 over denominator 2 straight pi end fraction fraction numerator straight A subscript straight gamma straight p subscript 0 over denominator straight V subscript 0 straight M end fraction
  • fraction numerator 1 over denominator 2 straight pi end fraction fraction numerator straight V subscript 0 Mp subscript 0 over denominator straight A squared straight gamma end fraction
  • fraction numerator 1 over denominator 2 straight pi end fraction square root of fraction numerator straight A squared γp subscript 0 over denominator MV subscript 0 end fraction end root
  • fraction numerator 1 over denominator 2 straight pi end fraction square root of fraction numerator straight A squared γp subscript 0 over denominator MV subscript 0 end fraction end root

C.

fraction numerator 1 over denominator 2 straight pi end fraction square root of fraction numerator straight A squared γp subscript 0 over denominator MV subscript 0 end fraction end root


FBD of piston at equilibrium
⇒ Patm A + mg = P0A

FBD of piston when piston is pushed down a distance x
straight P subscript atm space plus mg space minus space left parenthesis straight P subscript 0 plus dP right parenthesis space straight S space equals space straight m fraction numerator straight d squared straight x over denominator dt squared end fraction space left parenthesis ii right parenthesis
process space is space adiabatic space
rightwards double arrow space PV to the power of straight gamma space equals space straight C
rightwards double arrow straight C space equals dp equals space γPdV over straight V space space.. space left parenthesis iii right parenthesis
from space left parenthesis straight i right parenthesis space comma space left parenthesis ii right parenthesis space and space left parenthesis iii right parenthesis
straight f space equals space fraction numerator 1 over denominator 2 straight pi end fraction square root of fraction numerator straight A squared γP subscript 0 over denominator MV subscript 0 end fraction end root


FBD of piston at equilibrium
⇒ Patm A + mg = P0A

FBD of piston when piston is pushed down a distance x
straight P subscript atm space plus mg space minus space left parenthesis straight P subscript 0 plus dP right parenthesis space straight S space equals space straight m fraction numerator straight d squared straight x over denominator dt squared end fraction space left parenthesis ii right parenthesis
process space is space adiabatic space
rightwards double arrow space PV to the power of straight gamma space equals space straight C
rightwards double arrow straight C space equals dp equals space γPdV over straight V space space.. space left parenthesis iii right parenthesis
from space left parenthesis straight i right parenthesis space comma space left parenthesis ii right parenthesis space and space left parenthesis iii right parenthesis
straight f space equals space fraction numerator 1 over denominator 2 straight pi end fraction square root of fraction numerator straight A squared γP subscript 0 over denominator MV subscript 0 end fraction end root

401 Views

One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in the figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct statement:

  • The change in internal energy in the process AB is -350 R.

  • The change in internal energy in the process BC is -500 R.

  • The change in internal energy in the whole cyclic process is 250 R.

  • The change in internal energy in the whole cyclic process is 250 R.


D.

The change in internal energy in the whole cyclic process is 250 R.

According to first law of thermodynamics, 
(i) change in internal energy from A to B i.e, 

increment straight U subscript AB space equals space nC subscript straight V left parenthesis straight T subscript straight B minus straight T subscript straight A right parenthesis space equals space 1 space straight x space fraction numerator 5 straight R over denominator 2 end fraction left parenthesis 800 minus 400 right parenthesis space equals space 1000 straight R
increment straight U subscript BC space equals space nC subscript straight V left parenthesis straight T subscript straight C minus straight T subscript straight B right parenthesis space equals space 1 space straight x space fraction numerator 5 straight R over denominator 2 end fraction left parenthesis 600 minus 400 right parenthesis space equals space minus 500 straight R
increment straight U subscript Total space equals space 0
increment straight U subscript CA space equals space nC subscript straight V left parenthesis straight T subscript straight A minus straight T subscript straight C right parenthesis space equals space 1 space straight x space fraction numerator 5 straight R over denominator 2 end fraction left parenthesis 400 minus 600 right parenthesis space equals space minus 500 straight R space

According to first law of thermodynamics, 
(i) change in internal energy from A to B i.e, 

increment straight U subscript AB space equals space nC subscript straight V left parenthesis straight T subscript straight B minus straight T subscript straight A right parenthesis space equals space 1 space straight x space fraction numerator 5 straight R over denominator 2 end fraction left parenthesis 800 minus 400 right parenthesis space equals space 1000 straight R
increment straight U subscript BC space equals space nC subscript straight V left parenthesis straight T subscript straight C minus straight T subscript straight B right parenthesis space equals space 1 space straight x space fraction numerator 5 straight R over denominator 2 end fraction left parenthesis 600 minus 400 right parenthesis space equals space minus 500 straight R
increment straight U subscript Total space equals space 0
increment straight U subscript CA space equals space nC subscript straight V left parenthesis straight T subscript straight A minus straight T subscript straight C right parenthesis space equals space 1 space straight x space fraction numerator 5 straight R over denominator 2 end fraction left parenthesis 400 minus 600 right parenthesis space equals space minus 500 straight R space

573 Views

Helium gas goes through a cycle ABCDA (consisting of two isochoric and two isobaric lines) as shown in the figure. Efficiency of this cycle is nearly:(Assume the gas to be close to ideal gas)

  • 15.4%

  • 9.1%

  • 10.5%

  • 10.5%


A.

15.4%

The efficiency of a process is defined as the ratio of work done to energy supplied.
Here,straight eta space equals space fraction numerator increment straight W over denominator increment straight Q end fraction space equals space fraction numerator Area space under space straight p minus straight V space diagram over denominator increment straight Q subscript AB plus space increment straight Q subscript BC end fraction
Where Cp and Cv are two heat capacities (molar)
therefore space space fraction numerator straight p subscript straight o straight V subscript straight o over denominator nC subscript straight v increment straight T subscript 1 space plus nC subscript straight p space increment straight T subscript 2 end fraction
space equals fraction numerator straight p subscript straight o straight V subscript straight o over denominator begin display style 3 over 2 end style nR left parenthesis straight T subscript straight B minus straight T subscript straight A right parenthesis plus begin display style 5 over 2 end style nR left parenthesis straight T subscript straight C minus straight T subscript straight D right parenthesis end fraction
equals space fraction numerator straight p subscript straight o straight V subscript straight o over denominator begin display style 3 over 2 left parenthesis 2 straight p subscript straight o straight V subscript straight o minus straight p subscript straight o straight V subscript straight o right parenthesis plus 5 over 4 left parenthesis 4 straight p subscript straight o straight V subscript straight o minus 2 straight p subscript straight o straight V subscript straight o right parenthesis end style end fraction
equals space fraction numerator straight p subscript straight o straight V subscript straight o over denominator begin display style 3 over 2 straight p subscript straight o straight V subscript straight o plus 5 over 4 2 straight p subscript straight o straight V subscript straight o end style end fraction
equals space fraction numerator 1 over denominator 6.5 end fraction space equals space 15.4 percent sign

The efficiency of a process is defined as the ratio of work done to energy supplied.
Here,straight eta space equals space fraction numerator increment straight W over denominator increment straight Q end fraction space equals space fraction numerator Area space under space straight p minus straight V space diagram over denominator increment straight Q subscript AB plus space increment straight Q subscript BC end fraction
Where Cp and Cv are two heat capacities (molar)
therefore space space fraction numerator straight p subscript straight o straight V subscript straight o over denominator nC subscript straight v increment straight T subscript 1 space plus nC subscript straight p space increment straight T subscript 2 end fraction
space equals fraction numerator straight p subscript straight o straight V subscript straight o over denominator begin display style 3 over 2 end style nR left parenthesis straight T subscript straight B minus straight T subscript straight A right parenthesis plus begin display style 5 over 2 end style nR left parenthesis straight T subscript straight C minus straight T subscript straight D right parenthesis end fraction
equals space fraction numerator straight p subscript straight o straight V subscript straight o over denominator begin display style 3 over 2 left parenthesis 2 straight p subscript straight o straight V subscript straight o minus straight p subscript straight o straight V subscript straight o right parenthesis plus 5 over 4 left parenthesis 4 straight p subscript straight o straight V subscript straight o minus 2 straight p subscript straight o straight V subscript straight o right parenthesis end style end fraction
equals space fraction numerator straight p subscript straight o straight V subscript straight o over denominator begin display style 3 over 2 straight p subscript straight o straight V subscript straight o plus 5 over 4 2 straight p subscript straight o straight V subscript straight o end style end fraction
equals space fraction numerator 1 over denominator 6.5 end fraction space equals space 15.4 percent sign

1344 Views