Tag: heat and thermodynamics

Quizzes (20) Questions (162)

Questions Related to heat and thermodynamics

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).

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

For an isothermal expansion of an ideal gas, mark wrong statement

  1. there is no change in the temperature of the gas

  2. there is no change in the internal energy of the gas

  3. the work done by the gas is equal to the heat supplied to the gas

  4. the work done by the gas is equal to the change in its internal energy
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$In\quad isothermal\quad expansion.\ The\quad temperature\quad is\quad constant,\ so\quad the\quad change\quad in\quad temperature,\quad \triangle T\quad =\quad 0,\ as\quad \triangle v\quad =\quad \frac { 3 }{ 2 } \quad R& T\ and\quad \triangle T\quad =\quad 0\ \therefore \quad \quad \triangle V\quad =\quad 0\ \ \quad \quad \quad \quad \quad \quad \quad \triangle V\quad =\quad Q-W\ \quad \quad \quad \quad \quad \quad \quad \quad 0\quad =\quad Q-W\ \quad \quad \quad \quad \quad \quad \quad \quad Q\quad =\quad W\ Work\quad done\quad in\quad isothermal\quad process\quad =\quad nRln\frac { { V } _{ 2 } }{ { V } _{ 1 } } \ \quad \quad \quad nRln\frac { { V } _{ 2 } }{ { V } _{ 1 } } \quad \neq \quad 0\quad \ \quad \therefore \quad \triangle W\quad \neq \quad \triangle V\ Answer\quad :\quad D$

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

If a given mass of gas occupies a volume of 10 cc at 1 atmospheric pressure and temperature 100$^o$C. What will be its volume at 4 atmospheric pressure, the temperature being the same?

  1. 100 cc

  2. 400 cc

  3. 1.04 cc

  4. 2.5 cc
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

It is an isothermal process.
$P _1 V _1 = P _2 V _2$
$1 \times 10 = 4 \times V _2$
$V _2 = 2.5 cc$