Thermodynamics Test

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Thermodynamics Test - 43

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Question 1
1 / -0

If $$\Delta G$$ for a reaction is negative, the change is:

A
reversible

B
spontaneous

C
equilibrium

D
non-spontaneous

Solution

Solution:- (B) Spontaneous
For a spontaneous reaction,
$$\Delta{G} = -ve$$
Question 2
1 / -0

Determine $$ \Delta U^o $$ at 300 K for the following reaction using the listed enthalpies of reaction :
$$ 4CO(g)+8H_{ 2 }(g)\rightarrow 3CH_{ 4 }(g)+CO_{ 2 }(g)+2H_{ 2 }O(I) $$
$$ C (graphite)+1/2O_2(g) \rightarrow CO(g); \Delta H^o_1 = -110.5 kJ $$
$$ CO(g) +1/2 O_2(g) \rightarrow CO_2(g); \Delta H^o_2 = -282.9 kJ $$
$$ H_2(g) + 1/2 O_2(g) \rightarrow H_2O(I); \Delta H^o_3 = -285.8 kJ $$
$$ C(graphite)+2H_2(g) \rightarrow CH_4(g); \Delta H^o_4 = -74.8 kJ $$

A
$$-653.5 kJ$$

B
$$-686.2kJ$$

C
$$-747.4kJ$$

D
None of these

Solution
Question 3
1 / -0

The Normal boiling point of a liquid 'X' is $$400$$ K. Which of the following statement is true about the process $$X(l) \rightarrow X(g)$$?

A
At $$400$$ K and 1 atm pressure $$\Delta G=0$$

B
At $$400$$K and 2 atm pressure $$\Delta G=+ve$$

C
At $$400$$K and 0.1 atm pressure $$\Delta G=-'ve$$

D
At $$410$$K and 1 atm pressure $$\Delta G=+ve$$

Solution
Question 4
1 / -0

For hydrogenation of ethene into ethane as per following chemical reaction $$CH_2=CH_2+H_2 \xrightarrow[]{Pd} CH_3-CH_3$$
The $$\Delta H_{reaction}$$ is written as
Calculate $$\Delta H_{reaction}$$ for

A
$$\Delta H_{reaction}=4 \Delta H_{C-H}+ \Delta H_{C=C} + \Delta H_{H-H}-6 \Delta H_{H-H}-2 \Delta H_{C-C}$$

B
$$\Delta H_{reaction}=4 \Delta H_{C-H}+ 2\Delta H_{C-C} + \Delta H_{H-H}-6 \Delta H_{C-H}- \Delta H_{C-C}$$

C
$$\Delta H_{reaction}=4 \Delta H_{C-H}+ \Delta H_{C=C} + \Delta H_{H-H}-6 \Delta H_{C-H}- \Delta H_{C-C}$$

D
$$\Delta H_{reaction}= \Delta H_{C-H}+ \Delta H_{C=C} + \Delta H_{H-H}- \Delta H_{C-H}- \Delta H_{C-C}$$

Solution
Question 5
1 / -0

For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter $$\Delta U$$ and w correspond to

A
$$\Delta U<0,w=0$$

B
$$\Delta U<0,w<0$$

C
$$\Delta U>0,w=0$$

D
$$\Delta U>0,w>0$$

Solution

In bomb calorimeter, there is no expansion in volume, so, work done will be zero.

This reaction is exothermic. So, some heat will be evolved which will result in lowering of internal energy.

Hence, $△ U < 0$ & $w = 0$ .
Question 6
1 / -0

An ideal gas is taken through a cyclic thermo dynamical process through four steps. The amounts of heat involved in steps are
$$Q_1 = 5960 J, Q_2 = -5585 J, Q_3 = -2980 J, Q_4 = 3645 J$$; respectively, the corresponding works involved are $$W_1 = 2200 , W_2 = -825 J, W_3 = -1100 J$$ and $$W_4$$ respectively. Find the value of $$W_4$$ and efficiency of the cycle

A
1315 J, 10%

B
765 J, 11%

C
both of them

D
none of these

Solution

Given;-
$$\begin{matrix} { Q_{ 1 } }=5960J\, \, \, ,{ W_{ 1 } }=2200J \\ { Q_{ 2 } }=5585J\, \, \, ,{ W_{ 2 } }=825J \\ { Q_{ 3 } }=2980J\, \, \, ,{ W_{ 3 } }=1100J \\ { Q_{ 4 } }=3645J\, \, \, ,{ W_{ 4 } }=find \\ \eta =find \\ \end{matrix}$$
In cycic process $$\Delta U=0$$
$$A/q$$ $$1st$$ law of thermodynamics
$$\Delta Q= \Delta U+W$$
$$\Rightarrow +1040 = 275+W_4$$
$$\Rightarrow W_4=1040-275$$
$$=765$$
$$\eta \% = \dfrac{{output}}{{input}} \times 100$$

$$ = \dfrac{{1040}}{{5960 + 3645}}$$

$$ = \dfrac{{208}}{{1321}} \times 100$$
Hence, we get
$$\eta = 10.82\% $$
Question 7
1 / -0

Assertion : In Free expansion, $$\Delta U = 0$$
Reason : No work is done in free expansion

A
If both assertion and reason are true and reason is the correct explanation of assertion.

B
If both assertion and reason are true but reason is not the correct explanation of assertion.

C
If assertion is true but reason is false.

D
If both assertion and reason are false.

Solution

Both are true but unrelated.
Question 8
1 / -0

At $$5\times 10^{5}$$ bar pressure density of diamond and graphite are 3 g/cc adn 2 g/ cc respectively. at certain temperature 'T' Find the value of $$\Delta U-\Delta H$$ for the conversion of 1 mole of graphite to 1 mole of diamond at termperature 'T'

A
100 KJ/mol

B
50 KJ/mol

C
-100 KJ / mol

D
None of these

Solution
Question 9
1 / -0

$$5$$ mole of ideal gas at $$100K$$ ($${C}_{v,m}=28J/mol/K$$). It is heated upto $$200K$$. Calculate $$\Delta U$$ and $$\Delta (PV)$$ for the process. ($$R=8J/mol-K$$)

A
$$\Delta U=28kJ;\Delta (PV)=8KJ$$

B
$$\Delta U=14kJ;\Delta (PV)=4KJ$$

C
$$\Delta U=14kJ;\Delta (PV)=8KJ$$

D
$$\Delta U=28kJ;\Delta (PV)=4KJ$$

Solution

$$\Delta U=n{ C }_{ v }\Delta T=\cfrac { 5\times 28\times 100 }{ 1000 } =14KJ$$
$$\Delta (PV)=\cfrac { 5\times 8\times 100 }{ 1000 } =4KJ\quad $$
Question 10
1 / -0

A diatomic ideal gas is expanded at constant at pressure. If work done by the system is $$10 \,J$$ then calculate heat absorbed.

A
$$40 \,J$$

B
$$20 \,J$$

C
$$35 \,J$$

D
$$15 \,J$$

Solution

$$\Delta U = nCv_m \Delta T$$ $$W = -nR\Delta T$$
$$-10 = -nR\Delta T$$

$$\therefore \Delta U = n \times \dfrac{5}{2}R \times \dfrac{10}{nR}$$ or $$\Delta T = \dfrac{10}{nR}$$
or $$\Delta U = 25$$ $$\therefore \Delta U = q + W$$
$$25 = q - 10 \Rightarrow q = 35 \,J$$