Tag: heat and thermodynamics

Questions Related to heat and thermodynamics

Expansion of 1 mole of ideal gas is taking place from 2 litres to 8 litres at 
300 K against 1 atm pressure. calculate $ \Delta JK^{-1} mol^{-1} $ given $ R= \dfrac {8.3J}{mol-K}, 1 lit-atm= 100 j,in 2=0.693) $

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

  1. $11.5$

  2. $13.5$

  3. $9.5$

  4. $22.5$

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Correct Option: B

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 -

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

  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

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Correct Option: C

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

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

  1. 1 atm

  2. 2 atm

  3. 2.5 atm

  4. 3 atm

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Correct Option: A

In isobaric process of ideal gas $(f = 5)$ work done by gas is equal to $10\ J$. Then heat given to gas during process is

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

  1. $25\ J$

  2. $15\ J$

  3. $45\ J$

  4. $35\ J$

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Correct Option: D
Explanation:

$\triangle Q = \left (\dfrac {f}{2} + 1\right ) nR\triangle T = \dfrac {7}{2} \times 10 = 35\ J$.

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 $(R=2cal.mol-1K-1)$

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

  1. $1381.1cal$

  2. zero

  3. $163.7cal$

  4. $9lit.atm$

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Correct Option: B

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:

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

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

Show answer
Correct Option: A
Explanation:

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

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

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

  1. $-10368 J$

  2. $-110368 J$

  3. $12000 J$

  4. $120368 J$

Show answer
Correct Option: A
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, $\displaystyle{W = 2.3026 \times 3 \times 8.31 \times 300 \times log _{10}\frac{1}{4}}$
= $17221.15 (-2\log _{10} 2$)
= $-17221.15 \times 2 \times 0.3010 = -10368J$ 

A physical quantity that is conserved in a process

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

  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

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Correct Option: A

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

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

  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

Show answer
Correct Option: D
Explanation:

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

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?

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

  1. 100 cc

  2. 400 cc

  3. 1.04 cc

  4. 2.5 cc

Show answer
Correct Option: D
Explanation:

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