Tag: thermodynamic processes

Quizzes (0) Questions (39)

Questions Related to thermodynamic processes

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

The work done by 100 calorie of heat in isothermal expansion of ideal gas is 

  1. 418.4J

  2. 4.184J

  3. 41.84J

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

One calorie is equivalent to $4.184$ joules 

or,   $J=4.18 J/cal$.
Work done (W)=$ = 4.18 \times 500 = 2090\,J$
As $1$ calorie is equal to $4.184$ calorie then $100$ calorie is equal to $ = 4.184 \times 100 = 418.4$

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

The pressure and volume of a given mass of gas at a given temperature are $ \mathrm{P}  $ and $ \mathrm{V}  $ respectively.Keeping temperature constant, the pressure is increased by 10$  \%  $ and then decreased by 10$  \%  $ .The volume how will be -

  1. less than $ \mathrm{V} $

  2. more than $ \mathrm{V} $

  3. equal to $ \mathrm{V} $

  4. less than $ V $ for diatomic and more than $ V $ for monoatomic
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

Two bulbs of volume V and 4V contains gas at pressures of 5 atm,. 1 atm and at temperatures of 300K and 400K respectively. When these bulbs are joined by narrow tube keeping their temperature at their initial values. The pressure of the system is

  1. 1 atm

  2. 2 atm

  3. 2.5 atm

  4. 3 atm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the ideal gas law for the two bulbs: n1 = P1V1/RT1 and n2 = P2V2/RT2. Total moles n = n1 + n2. Final pressure P = nRT_final / V_total. Since temperatures are kept constant, we use the weighted average pressure approach.

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

One mole of an ideal gas at $300K$ is expanded isothermally from an initial volume of $1litre$ to $10litres$. The $\Delta E$ for this process is $(R=2cal.mol-1K-1)$

  1. $1381.1cal$

  2. zero

  3. $163.7cal$

  4. $9lit.atm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For an ideal gas, internal energy E is a function of temperature only (E = f(T)). In an isothermal process, deltaT = 0, therefore deltaE = 0.

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

An Ideal gas undergoes an isobaric process. If its heat capacity is $C _v$ at constant volume and number of mole $n$. then the ratio of work done by gas to heat given to gas when temperature of gas changes by $\Delta T$ is:

  1. $\left(\dfrac{nR}{c _v + R}\right)$

  2. $\left(\dfrac{R}{c _v + R}\right)$

  3. $\left(\dfrac{nR}{c _v - R}\right)$

  4. $\left(\dfrac{R}{c _v - R}\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac{f}{2} R = \dfrac{C _v}{n}$
$W = nR \Delta T$
$\Delta Q = \left(\dfrac{f}{2} + 1\right) nR \, \Delta T$
$\dfrac{W}{\Delta Q} = \left(\dfrac{2}{f + 2} \right) = \dfrac{2}{\dfrac{2C _v}{nR} + 2} = \left(\dfrac{nR}{C _v + R} \right)$

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

Three moles of an ideal gas kept at a constant temperature at $300 K$ are compressed from a volume of $4 L$ to $1 L$. The work done in the process is

  1. $-10368 J$

  2. $-110368 J$

  3. $12000 J$

  4. $120368 J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Work done in an isothermal process is given by 
$\displaystyle{W = 2.3026nRT\log _{10}\frac{V _2}{V _1}}$
Here, $n = 3, R = 8.31 J/mol^oC$
$T = 300 K$, $V _1= 4 L$, $V _2 = 1 L$
Hence, $\displaystyle{W = 2.3026 \times 3 \times 8.31 \times 300 \times log _{10}\frac{1}{4}}$
= $17221.15 (-2\log _{10} 2$)
= $-17221.15 \times 2 \times 0.3010 = -10368J$ 

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

A physical quantity that is conserved in a process

  1. must have the same value for all observers

  2. can never take negative values

  3. must be dimensionless

  4. need not necessarily be a scalar
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A conserved quantity in physics, such as energy or momentum, is independent of the observer's frame of reference (though its specific value might change, the property of being conserved is invariant).