Thermodynamics Test

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Thermodynamics Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

The molar heat capacities of Iodine vapour and solid are $$7.8$$ and $$14\ cal/mol$$ respective enthalpy of sublimation of iodine is $$6096\ cal/mol$$ at $$200^{\circ}C$$, then what is $$\Delta H$$ the value at $$250^{\circ}C$$ in cal/mol.

A
$$5360$$

B
$$4740$$

C
$$6406$$

D
none of these

Solution

$$\Delta C_p = \dfrac{\Delta H_2 - \Delta H_1}{T_2-T_1}$$
$$ 14-7.8 = \Delta {H_2 - 6096}{50}$$
$$6.2\times 50 = \Delta H_2 - 6096$$
$$6.2 \times 50 + 6096 = \Delta H_2$$
$$\Delta H_2 = 310+ 6096 = 6406 $$ cal/mol
Question 2
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$$100\ ml\ 0.5 N\ H_{2}SO_{4}$$ (strong acid) is neutralised with $$200\ ml\ 0.2\ M\ NH_{4}OH$$ in a constant pressure Calorimeter which results in temperature rise of $$1.4^{\circ}C$$. If heat capacity of Calorimeter content is $$1.5\ kJ/^{\circ}C$$. Which statement is/ are correct:

Given: $$HCl + NaOH\rightarrow NaCl + H_{2}O + 57\ kJ$$
$$CH_{3}COOH + NH_{4}OH \rightarrow CH_{3}COONH_{4} + H_{2}O + 48.1\ kJ$$

A
enthalpy of neutralisation of $$HCl$$ v/s $$NH_{4}OH$$ is $$-52.5\ kJ/mol$$

B
enthalpy of dissociation (ionization) of $$NH_{4}OH$$ is $$4.5\ kJ/ mol$$

C
enthalpy of dissociation of $$CH_{3}COOH$$ is $$4.6\ kJ/mol$$

D
$$\triangle H$$ for $$2H_{2}O(l)\rightarrow 2H^{+}(aq.) + 2OH^{-}(aq.)$$ is $$114\ kJ$$

Solution
Question 3
1 / -0

Fruend and Long were 2 scientists interested in adsorption.
Once in a discussion Fruend asked "I can see adsorption is spontaneous, but why it is always exothermic?"
Long said"_________________"
Choose Long's answer :

A
$$\triangle$$G > 0 & $$\triangle$$ S < 0 $$\Rightarrow$$ $$\triangle$$H < 0

B
$$\triangle$$G < 0 & $$\triangle$$ S < 0 $$\Rightarrow$$ $$\triangle$$H < 0

C
$$\triangle$$G > 0 & $$\triangle$$ S > 0 $$\Rightarrow$$ $$\triangle$$H > 0

D
$$\triangle$$G , 0 & $$\triangle$$ S > 0 $$\Rightarrow$$ $$\triangle$$H > 0

Solution

For the adsorption-
$$\Delta G, \Delta H , \Delta S$$ all are negative.
According to Gibbs free energy-
$$\Delta G = \Delta H - T\Delta S$$
$$\Delta G =\Delta H +T\Delta S$$, as the $$\Delta S = -ve$$
So $$\Delta H = \Delta G- T\Delta S$$
As the $$\Delta G$$ is already negative so $$\Delta H$$ also comes negative.
Question 4
1 / -0

For the reaction, $${ C }_{ 3 }{ H }_{ 8 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 3C{ O }_{ 2 }\left( g \right) +4{ H }_{ 2 }O\left( l \right) $$ at constant temperature, $$\Delta H-\Delta E$$ is:

A
$$+3 RT$$

B
$$-RT$$

C
$$+RT$$

D
$$-3RT$$

Solution

$${ C }_{ 3 }{ H }_{ 8 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 3C{ O }_{ 2 }\left( g \right) +4{ H }_{ 2 }O\left( l \right) $$

$$ \displaystyle \Delta n $$ is the difference in the number of moles of gaseous products and gaseous reactants.

$$ \displaystyle \Delta n = (3)-(1+5)=-3$$

$$ \displaystyle \Delta H=\Delta E + \Delta nRT$$

$$ \displaystyle \Delta H-\Delta E = \Delta nRT$$

$$ \displaystyle \Delta H-\Delta E = -3RT$$

Hence.option D is correct.
Question 5
1 / -0

$$2$$ moles of $$CO$$ and $$1$$ mole of $$O_{2}$$ are taken in a container of volume $$1$$ litre to form $$2$$ moles of $$CO_{2}$$ according to equation

$$2CO + O_{2}\rightarrow 2CO_{2}, \ \Delta H = -560KJ$$.

In this process, pressure changes from $$70\ atm$$ to $$40\ atm$$. Find out value of $$\Delta U$$ at $$500\ K$$.

[Given: gases deviates apprecially from ideal nature. $$1\ atm-litre = 0.1\ KJ$$].

A
$$-530\ KJ$$

B
$$-500\ KJ$$

C
$$-557\ KJ$$

D
$$-560\ KJ$$

Solution
Question 6
1 / -0

A 10 g piece of iron $$(C=0.45 \,J/g^0 C)$$ at $$100^0 C$$ is dropped in to $$25 g$$ of water $$(C=4.2\,J/g^0 C)$$ at $$ 27^0 C$$ . Find temperature of the iron and water system at thermal equilibrium.

A
$$30^0 C$$

B
$$27^0 C$$

C
$$40^0 C$$

D
None of these

Solution

Let T be temperature at equilibrium,
According to the principle of calorimetry,
Heat gained by water = Heat lost by iron
$$\Rightarrow mc\Delta T = mc\Delta T$$
$$\Rightarrow 25 × 4.2 ×(T- 27)= 10×0.45 ×(T-100)$$
$$\Rightarrow 105T - 2835 = 4.5T - 450$$
$$\Rightarrow 100.5 T= 2385$$
$$\Rightarrow T= 23.7°C$$
Hence, the correct option is D.
Question 7
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Which statement is correct?

A
$$\left (\dfrac {dH}{dT}\right )_{P} < \left (\dfrac {dE}{dT}\right )_{V}$$

B
$$\left (\dfrac {dH}{dT}\right )_{P} + \left (\dfrac {dE}{dT}\right )_{V} = R$$

C
$$\left (\dfrac {dE}{dV}\right )_{T}$$ of ideal gas is zero

D
All of these

Solution
Question 8
1 / -0

Calculate the hear needed to raise the temperature of $$20g$$ from $${25}^{o}C$$ to $${500}^{o}C$$, if specific heat capacity of iron is $$0.45J{K}^{-1}$$ $${g}^{-1}$$.

A
$$4274J$$

B
$$225J$$

C
$$15.66J$$

D
$$2250J$$

Solution
Question 9
1 / -0

A solid material supplied with heat at a constant rate. The temperature of material is changing with heat input as shown in figure. What does slope DE represents?

A
latent heat of liquid

B
latent heat of vapour

C
heat capacity of vapour

D
inverse of heat capacity of vapour

Solution
Question 10
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$$1$$ mole of ice at $$0^{\circ}C$$ and $$4.6$$ mm Hg pressure is converted to water vapour at a constant temperature and pressure. Find $$\Delta H$$ if the latent heat of fusion of ice is $$80$$ Cal/gm and latent heat of vaporization of liquid water at $$0^{\circ}C$$ is $$596$$ Cal per gram and the volume of ice in comparison to that of water (vapour) is neglected.

A
14.53 Kcal/mol

B
12.16 Kcal/mol

C
10.22 Kcal/mol

D
15.62 Kcal/mol

Solution

$$\text{ice} \longrightarrow\text{vapour}$$

Given that:-

$$\Delta{{H}_{f}} = 80 \; {cal}/{gm} = 80 \times 18 = 1.44 \; {kcal}/{mol}$$

$$\Delta{{H}_{v}} = 596 \; {cal}/{gm} = 596 \times 18 = 10.728 \; {Kcal}/{mol}$$

$$\Delta{H} = \Delta{{H}_{f}} + \Delta{{H}_{v}}$$

$$\Delta{H} = 1.44 + 10.728 = 12.168 \; {Kcal}/{mol}$$

Hence, the correct option is $$\text{B}$$