Tag: introduction to thermodynamics

Questions Related to introduction to thermodynamics

Which of the following state function not zero at standard state?

  1. Enthalpy

  2. Entropy

  3. Free energy

  4. None of the above


Correct Option: B,C
Explanation:

Solution:- (B) entropy and (C) free energy

Entropy and free energy are the state function which are not zero at standard state.

Work is __________ function.

  1. path

  2. state 

  3. both path and state

  4. none of these


Correct Option: A
Explanation:

Work is a path function and not a state function. The amount of work done is different for different paths although the initial and final states are the same. The amount of work depends on the path connecting the initial and final states.

Which of the following is/are not state function?

  1. $q$

  2. $q-w$

  3. $\cfrac { q }{ w } $

  4. $q+w$


Correct Option: A,B,C

A system absorbs 186 kJ of heat and the surroundings do 120 kJ of work on the system. What is the change in internal energy of the system? Express the internal energy in kilojoules to three significant figures.

  1. $440KJ$

  2. $360KJ$

  3. $120.32J$

  4. $-200J$


Correct Option: B

The work done in an open vessel at $300$K, when $112g$ iron reacts with dilute $HCl$ to give $FeCl _2$, is nearly:

  1. $1.1$ kcal

  2. $0.6$ kcal

  3. $0.3$ kcal

  4. $0.2$ kcal


Correct Option: A
Explanation:
The reaction involved is:
$Fe+2HCl \rightarrow FeCl _2+H _2$
Atomic mass of $Fe=56 g/mol $
Thus, 56 g of Iron reacts with 2 moles of HCl to give one mole of Hydrogen gas.
Initial volume of $H _2$ gas = $V _1 = 0$
Final volume of $H _2$ gas=$ V _2$
Using ideal gas law:$PV = n R T$
where n=mass/molar mass, R=8314 J/K/mol and given that T=300 K
$PV _2 = (112/56) \times 8.314 \times 300 = 4988.4J$
Work done $= -P \Delta V=-P(V _2 - V _1) =-P V _2 = -4988.4J$
negative work done is work of expansion.
since 4184 J=1 kcal
thus $4988.4 J=1.19 kcal$ of work is done by the system.

An open vessel at $27^{\circ}C$ is heated until ($\tfrac 25$)th of the air in it has been expelled. Assuming that the volume of the vessel remains constant. Which of the following option is correct?

  1. Final temperature is 500 K

  2. Final volume is 2 times the initial volume

  3. Final pressure is 1 atm

  4. None of the above


Correct Option: A
Explanation:

$T _{1}=27°C=27+273=300 K$


Let $n _{1}$ be n.

$n _{2}=n-\cfrac{2n}{5}=\cfrac{3n}{5}$

$T _{2}$=?

We know that, $n _{1}T _{1}=n _{2}T _{2}$

$\Longrightarrow n\times 300=\cfrac { 3n }{ 5 } \times { T } _{ 2 } \ \Longrightarrow { T } _{ 2 }=500K=500-273=227°C$

An open vessel containing air is heated from 300 K to 400 K. The fraction of air originally present which goes out of it is:

  1. 3/4

  2. 1/4

  3. 2/3

  4. 1/8


Correct Option: B
Explanation:
 
According to ideal gas equation
$PV = nRT$
$\Rightarrow \; n \propto \cfrac{1}{T}$
$\Rightarrow \; \cfrac{{n} _{1}}{{n} _{2}} = \cfrac{{T} _{2}}{{T} _{1}}$
Given that:-
${n} _{1} = 1$ 
${T} _{1} = 300K$, 
${T} _{2} = 400K$ 
${n} _{2} = ?$
$\therefore \; \cfrac{1}{{n} _{2}} = \cfrac{400}{300}$
${n} _{2} = \cfrac{3}{4}$
The fraction of air present in the vessel after heating ${n} _{2} = \cfrac{3}{4}$
The fraction of air which goes out of the vessel $= 1 - \cfrac{3}{4} = \cfrac{1}{4}$

System in which there is no exchange of matter, work or energy from surroundings is:

  1. closed

  2. adiabatic

  3. isolated

  4. isothermal


Correct Option: C
Explanation:

Closed System - The system in which only energy can be exchanged with the surrounding.


Adiabatic - The process in which heat is not exchanged by the system with the surroundings, i.e., $Q=0$.


Isolated - The system in which neither energy nor matter can be exchanged with the surroundings.  

Isothermal - The temperature remains constant, i.e., $\Delta T=0$.


Hence, the correct answer is option $\text{C}$.

In a closed system : $A\left( s \right) \rightleftharpoons 2B\left( g \right) +3C\left( g \right) $ if the partial pressure C is of doubled then partial pressure B wil be:

  1. Twice the original pressure

  2. Half of its original pressure

  3. $\dfrac { 1 }{ 2\sqrt { 2 } } $ times, the original pressure

  4. $2\sqrt { 2 } $ times its original pressure


Correct Option: C
Explanation:
Solution:- (C) $\cfrac{1}{2 \sqrt{2}}$ times, the original pressure
${A} _{\left( s \right)} \rightleftharpoons 2 {B} _{\left( g \right)} + 3 {C} _{\left( g \right)}$
${K} _{P} = {\left( {P} _{B} \right)}^{2} {\left( {P} _{C} \right)}^{3} ..... \left( 1 \right)$
If we double the partial pressure of $C$, i.e., ${P} _{C}' = 2 {P} _{C}$
$\therefore {K} _{P}' = {\left( {P} _{B}' \right)}^{2} {\left( {P} _{C}' \right)}^{3}$
$\Rightarrow {K} _{P}' = {\left( {P} _{B}' \right)}^{2} {\left( 2 {P} _{C} \right)}^{3}$
$\Rightarrow {K} _{P}' = 8 {\left( {P} _{B}' \right)}^{2} {\left( {P} _{C} \right)}^{3}$
Since ${K} _{P}$ is constant,
$\therefore {K} _{P} = {K} _{P}'$
$\Rightarrow {\left( {P} _{B} \right)}^{2} {\left( {P} _{C} \right)}^{3} = 8 {\left( {P} _{B}' \right)}^{2} {\left( {P} _{C} \right)}^{3}$
$\Rightarrow {P} _{B}' = \sqrt{\cfrac{{P} _{B}}{8}}$
$\Rightarrow {P} _{B}' = \cfrac{{P} _{B}}{2 \sqrt{2}}$
Hence the partial pressure of $B$ will be $\cfrac{1}{2 \sqrt{2}}$ times of its original pressure.

Which of the following statement is correct?

  1. The presence of reaction species in a covered beaker is an example of open system.

  2. There is an exchange of energy as well as matter between the system and the surroundings in a closed system.

  3. The presence of reactants in a closed vessel made up of copper is an example of a closed system.

  4. The presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system.


Correct Option: C
Explanation:

In a closed system (For example, the presence of reactants in a closed vessel made of conducting material e.g copper) there is no exchange of matter, but exchange of energy is possible between system and the surroundings.