Tag: chemical thermodynamics

Questions Related to chemical thermodynamics

Which of the following is path function?

chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

  1. Work

  2. Specific volume

  3. Pressure

  4. Temperature

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Correct Option: A
Explanation:

In thermodynamics, a state function or function of state is a function defined for a system relating several state variables or state quantities that depends only on the current equilibrium state of the system. State functions do not depend on the path by which the system arrived at its present state. A state function describes the equilibrium state of a system.

Hence, work is a path function and not state function.

An isolated system is that system in which:

chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

  1. there is no exchange of energy with the surroundings

  2. there is exchange of mass and energy with the surroundings

  3. there is no exchange of energy and mass with the surroundings

  4. there is exchange of energy and mass with the surroundings

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Correct Option: C
Explanation:

An isolated system is a thermodynamic system that cannot exchange either energy or matter (mass) outside the boundaries of the system. An isolated system differs from a closed system by the transfer of energy. Closed systems are only closed to matter, energy can be exchanged across the system's boundaries.
Hence, correct answer is option C. 

Steady state is represented by :

chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

  1. Getting raw materials.

  2. Intake of food and energy.

  3. Intake of materials and energy, elimination of wastes and dissipation of energy.

  4. Removal of waste products and intake of raw materials.

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Correct Option: C
Explanation:
Living beings require energy for their day to day activity. They generally consume food for intake of energy. They generally consume food for intake of energy, The utilize energy and eliminate the rest of the waste.
In the whole process, there is no change in energy and mass. Hence it can be considered as a steady state.
To get clear understanding of steady state read the following example:
When the drain of bathtub is open and water is being added into the bathtub at same rate, then the amount of water in the bathtub is constant.
Water draining=Water added
It is a steady state as the state variable volume is constant overtime.

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

chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

  1. Enthalpy

  2. Entropy

  3. Free energy

  4. None of the above

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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.

chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

  1. path

  2. state 

  3. both path and state

  4. none of these

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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?

chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

  1. $q$

  2. $q-w$

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

  4. $q+w$

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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.

chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

  1. $440KJ$

  2. $360KJ$

  3. $120.32J$

  4. $-200J$

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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:

chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

  1. $1.1$ kcal

  2. $0.6$ kcal

  3. $0.3$ kcal

  4. $0.2$ kcal

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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?

chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

  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

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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:

chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

  1. 3/4

  2. 1/4

  3. 2/3

  4. 1/8

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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 $= 1 - \cfrac{3}{4} = \cfrac{1}{4}$