Tag: measurement and effects of heat

Questions Related to measurement and effects of heat

At constant pressure how much fraction of heat supplied to gas is converted into mechanical work ?  

  1. $\dfrac { \gamma -1 }{ \gamma } $

  2. $\dfrac { \gamma }{ \gamma -1 } $

  3. $\gamma -1$

  4. $\dfrac { \gamma }{ \gamma +1 } $


Correct Option: A

A gas compressed to half of its volume at ${30}^{o}C$. Upto what temperature should it be heated, so that its volume increase to double of its original volume?

  1. ${60}^{o}C$

  2. $303K$

  3. $606K$

  4. $1212K$


Correct Option: C

An ideal gas is expanding,such that $P T^2= constant$.The coefficient of volume expansion of the gas is 

  1. $1/T$

  2. $2/T$

  3. $3/T$

  4. $4/T$


Correct Option: D

A gas follows $VT^2 =$ const. Its volume expansion coefficient will be :-

  1. $\dfrac{2}{T}$

  2. $-\dfrac{2}{T}$

  3. $\dfrac{3}{T}$

  4. $-\dfrac{3}{T}$


Correct Option: C
Explanation:

$\begin{array}{l} \gamma =\dfrac { 1 }{ V } \left( { \dfrac { { dV } }{ { dT } }  } \right)  \ P{ T^{ 2 } }=cons\tan  t \ \dfrac { { nRT } }{ V } { T^{ 2 } }=cons\tan  t \ \therefore \gamma =\dfrac { 3 }{ T }  \ Hence,\, C\, the\, \, correct\, option\, .\,  \end{array}$

A one litre flask contain some mercury. It is found that at different temperatures the volume of air inside the flask remain same. the volume of mercury taken in the flask is (coefficient of linear expansion of volume expansion of $Hg$ is $1.8\times { 10 }^{ -4 }/ _{  }^{ o }{ C }$ 

  1. $150ml$

  2. $750ml$

  3. $1000ml$

  4. $700ml$


Correct Option: A

A uniform steel rod has length $\ell$ at $0^oC$. Now one of its end is kept in ice $(0^oC)$ and the other end is kept in steam $(100^oC)$. If the coefficient of thermal expansion of the rod is $\alpha,$how much is the thermal expansion of the rod at steady state? 

  1. $50\ \alpha\ell$

  2. $100\ \alpha\ell$

  3. $200\ \alpha\ell$

  4. $150\ \alpha\ell$


Correct Option: A

An inflated rubber balloon contains one mole of an ideal gas. Has a pressure p, volume V and temperature T. if the temperature rises to 1.1 T, and the volume is increase to 1.05 V, the final pressure will be:

  1. 1.04p

  2. 1.2 p

  3. less than p

  4. between p and 1.1.


Correct Option: A
Explanation:

$PV=nRT\rightarrow (1)\ P _1(1.05V)=nR(1.1T)\rightarrow (2)\ \Rightarrow (1)\div(2)\ \Rightarrow \cfrac{P}{1.05P _1}=\cfrac{1}{1.1}\ \Rightarrow P _1=\cfrac{1.1P}{1.05}=1.047P$

The pressures of a gas in the bulb of constant volume gas thermometer at 0$^{0}$ C are 54.6 cm and 74.6cm of Hg respectively. The pressure at 50$^{0}$ C is:

  1. 64.6cm of Hg

  2. 60.6cm of Hg

  3. 58.6cm of Hg

  4. 52.6cm of Hg


Correct Option: A
Explanation:
For a constant volume thermometer,
$P\propto T$ , the slope is given by
$\beta =\dfrac { { P } _{ 100 }-{ P } _{ 0 } }{ 100-0 }  $
$\beta =\dfrac { 74.6-54.6 }{ 100 }  $
$\beta$=0.2.
Now Applying equation for straight line,
${ P } _{ 50 }=T\beta +{ P } _{ 0 }$ 
${ P } _{ 50 }=50\times 0.2+54.6$
${ P } _{ 50 }=64.6$cm of Hg

A vessel contains 1 mole of an ideal monoatomic gas. The coefficient of volume expansion of the gas is $\alpha $. 2 moles of a diatmoic; ideal gas is then introduced into the same vessel. The coefficient of the volume expansion of the mixture will be

  1. $3\alpha /2$

  2. $2\alpha /3$

  3. $\alpha $

  4. $\alpha /3$


Correct Option: C
Explanation:

The volume coefficient of gas is given by,
${ \alpha  } _{ V }={ \left( \frac { 1 }{ V } \frac { \partial V }{ \partial T }  \right)  } _{ p }$
From the above equation it can be seen that it is independent of the number of moles,

If at $60^\circ$C and 80 cm of mercury pressure, a definite mass of a gas is compressed slowly, then the final pressure of the gas if the final volume is half of the initial volume $ (\gamma = \dfrac { 3 }{ 2 }$) is:

  1. 120 cm of Hg

  2. 140 cm of Hg

  3. 160 cm of Hg

  4. 180 cm of Hg


Correct Option: C
Explanation:
Given initial pressure, $P _1=80\,cm\,of\,Hg$

If the gas is compressed slowly, then the process is isothermal.

At constant temperature,

$P _1V _1=P _2V _2$

Given, 

Final volume is half of the initial volume.

That is, $V _2=\dfrac{V _1}{2}$

Final pressure, $P _2=\dfrac{P _1V _1}{V _2}=\dfrac{80 \times V _1}{V _1/2}=160\,cm\,of\,Hg$