Tag: heat and thermodynamics

Questions Related to heat and thermodynamics

A diatomic gas which has initial volume of $10$ litre is isothermally compressed to $1/15^{th}$ of its original volume where initial pressure is $10^5$ Pascal. If temperature is $27^o$C then find the work done by gas.

  1. $-2.70\times 10^3$J

  2. $2.70\times 10^3$J

  3. $-1.35\times 10^3$J

  4. $1.35\times 10^3$J


Correct Option: A
Explanation:

$w=nRT ln\left(\dfrac{v _2}{v _1}\right)$
$w=P _0V _0ln\left(\dfrac{v _2}{v _1}\right)$
$w=10^5\times 10\times 10^{-3}ln\left(\dfrac{1}{15}\right)$
$w=-2.70\times 10^3J$.

One mole of an ideal gas undergoes an isothermal change at temperature T so that its volume V is doubled. R is the molar gas constant. Work done by the gas during this change is :

  1. RT $\ln 4$

  2. RT $\ln 3$

  3. RT $ \ln 2$

  4. RT $ \ln 1$


Correct Option: C
Explanation:

Under isothermal process work is given by the relation $W = RT \ln (\dfrac{V _{f}}{V _{i}})$
Therefore work will be $W = RT \ln(2)$

The slope of adiabatic curve is ________ than the slope of an isothermal curve.

  1. Greater.

  2. lesser

  3. data insufficient

  4. can be both a and b


Correct Option: A

Three moles of an ideal gas $\left (C _{P} = \dfrac {7R}{2}\right )$ at pressure $P _{A}$ and temperature $T _{A}$ is isothermally expanded to twice the initial volume. The gas is then compressed at constant pressure to its original volume. Finally the gas is heated at constant volume to its original pressure $P _{A}$.
Calculate the net work done by the gas and the net heat supplied to the gas during the complete process.

  1. $0.579\ RT _{A}, \triangle Q = 0.579\ RT _{A}$.

  2. $79\ RT _{A}, \triangle Q = 0.679\ RT _{A}$.

  3. $0.9\ RT _{A}, \triangle Q = 0.779\ RT _{A}$.

  4. $0.7\ RT _{A}, \triangle Q = 0.979\ RT _{A}$.


Correct Option: A

Two soap bubbles having radii $3\ cm$ and $4\ cm$ in vacuum, coalesce under isothermal conditions. The radius of the new bubble is

  1. $1\ cm$

  2. $5\ cm$

  3. $7\ cm$

  4. $3.5\ cm$


Correct Option: B

One mole of an ideal gas u ndergoes a process:
$P = \dfrac{P _0}{1+(V _0/ V)^2}$.
Here $P _0$ and $V _0$ are constants. change in temperature of the gas when volume is changed from $V=V _0$ to $V = 2V _0$ is: 

  1. $-\dfrac{2P _0V _0}{5R}$

  2. $\dfrac{11P _0V _0}{10R}$

  3. $-\dfrac{5P _0V _0}{4R}$

  4. $P _0V _0$


Correct Option: B

Let $Q$ and $W$ denote the amount of heat given to an ideal gas and the work done by it in an isothermal process.

  1. $Q = 0$

  2. $W = 0$

  3. $Q \neq W$

  4. $Q = W$


Correct Option: D

Work done during isothermal expansion of one mole of an ideal gas $10$ atm to $1$ atm at $300\ K$ is

  1. $-4938.8\ J$

  2. $4938.8\ J$

  3. $-5744\ J$

  4. $6257.2\ J$


Correct Option: C

Identical cylinders contain helium at 2.5 atm and agron at 1 atm respectively. If the are filled  in one of the cylinder,the pressure would be.

  1. 3.6 atm

  2. 1.75 atm

  3. 1.5 atm

  4. 1.0 atm


Correct Option: B
Explanation:

Given that,

Pressure ${{p} _{1}}=2.5\,atm$

Pressure ${{P} _{2}}=1\,atm$

Volume ${{V} _{1}}={{V} _{2}}=V$

Both the cylinders are similar, volume of both the gases is equal

Let the volume of gases be V

The amount of pressure P in one of the cylinder will be equal to the total pressure at equilibrium

Now,

  $ P=\dfrac{{{P} _{1}}{{V} _{1}}+{{P} _{2}}{{V} _{2}}}{{{V} _{1}}+{{V} _{2}}} $

 $ P=\dfrac{2.5\times V+1\times V}{2V} $

 $ P=1.75\ atm $

Hence, the pressure is $1.75\ atm$

One mole of an ideal monoatomic gas is at $360K$ and a pressure of $10 ^ { 5 } Pa.$ It is compressed at constant pressure until its volume is halved. Taking $R$ as $8.3 J{ mol } ^ { - 1 }{ K } ^ { - 1 }$ and the initial volume of the gas as $3.0 \times 10 ^ { - 2 } { m } ^ { 3 }$ , the work done on the gas is

  1. $-1500 J$

  2. $+1500 J$

  3. $-3000 J$

  4. $+3000 J$


Correct Option: A