Chemistry Test

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Chemistry Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

Identify the correct statement regarding a spontaneous process.

A
Lowering of energy in the process is the only criterion for spontaneity.

B
For a spontaneous process in an isolated system, the change in entropy is positive.

C
Endothermic process are never spontaneous.

D
Exothermic process are always spontaneous.

Solution

In an isolated system, energy remains constant.
\(\therefore\) Driving force for a spontaneous process will be to increase entropy. Therefore, change in entropy is positive. \(\Delta G=\Delta H-T \Delta S\)
\(\Delta H=O\) for isolated system. \(-\Delta G=T \Delta S\)
\(\Delta GO\) for spontaneous process.
Question 2
1 / -0

A mono-atomic ideal gas undergoes a process in which the ratio of P to V at any instant is constant and is equal to 1. What is the molar heat capacity of the gas?

A
2R

B
3R/2

C
5R/2

D
0

Solution

Given that \(P\) is equal to \(V\). For one mole of an ideal gas, \(P V=R T\) \(P d V+V d P=R d T\) or \(2 P d V=R d T(\) as \(P=V)\)
or, \(P d V=\frac{R d T}{2} \ldots(1)\)
Now, from first law of thermodynamics, \(d q=d U+P d V\)
\(d q=C_{v} d T+P d V\)
\(d q=C_{v} d T+\frac{R d T}{2}(\) from equation 1\()\)
For ideal monoatomic gas, \(C_{v}=\frac{3 R}{2}\) Integrating, we get \(\frac{d q}{d T}=\frac{3 R}{2}+\frac{R}{2}=\frac{4 R}{2}=2 R\)
Question 3
1 / -0

The species which by definition has zero standard molar enthalpy of formation at 298 K is

A
Br₂(g)

B
Cl₂(g)

C
H₂O(g)

D
CH₄(g)

Solution

Bromine and water exist in liquid state at 298 K. Methane is not an elemental species. Cl₂ is a gas at 298 K.
Question 4
1 / -0

2 mol of an ideal gas expanded isothermally and reversibly from 1 litre to 10 litres at 300 K. What is the enthalpy change?

A
4.98 kJ

B
11.47 kJ

C
-11.47 kJ

D
0kJ

Solution

\(\mathrm{H}=\mathrm{E}+\mathrm{PV}\)
\(\Delta \mathrm{H}=\Delta \mathrm{E}+\Delta(\mathrm{PV})\)
or \(\Delta \mathrm{H}=\Delta \mathrm{E}+\mathrm{nR} \Delta \mathrm{T}\)
\(\Delta \mathrm{T}=0\)
\(\Delta \mathrm{E}=0\)
\(\therefore \Delta \mathrm{H}=0\)
Question 5
1 / -0

The internal energy change in the conversion of 2.0 moles of calcite form of CaCO₃ to the aragonite form is 0.4 kJ. Calculate the enthalpy change when the pressure is 1 bar. Given that the densities of solids are 2.7gcm⁻³ and 2.9gcm⁻³ respectively.

A
0.2 kJ/mol

B
2.0 kJ/mol

C
1.0 kJ/mol

D
10 kJ/mol

Solution

Calcite and aragonite are two forms of calcium carbonate. \(\Delta H\) for 1 mole of calcite to aragonite is \(\frac{0.4}{2}=0.2 k J\) \(\Delta H=\Delta E+P \Delta V\)
\(\Delta E=0.2 k J=0.2 \times 10^{3} J m o l^{-1}\)
\(P=1 b a r=1.0 \times 10^{5} P a\)
\(\Delta V=V_{\text {aragonite }}-V_{\text {calcite }}\) \(=\frac{100}{2.9}-\frac{100}{2.7}=-2.55 \mathrm{cm}^{3}=-2.55 \times 10^{-6} \mathrm{m}^{3}\)
\(\Delta H=\left(0.2 \times 10^{3}\right)-\left(1 \times 10^{5}\right) \times 2.55 \times 10^{-6}\)
\(=199.745 J=0.1997 k J\) for 1 mole
Question 6
1 / -0

Identify the correct statement regarding a spontaneous process.

A
Lowering of energy in the process is the only criterion for spontaneity.

B
For a spontaneous process in an isolated system, the change in entropy is positive.

C
Endothermic process are never spontaneous.

D
Exothermic process are always spontaneous.

Solution

In an isolated system, energy remains constant.
\(\therefore\) Driving force for a spontaneous process will be to increase entropy. Therefore, change in entropy is positive. \(\Delta G=\Delta H-T \Delta S\)
\(\Delta H=O\) for isolated system. \(-\Delta G=T \Delta S\)
\(\Delta GO\) for spontaneous process.
Question 7
1 / -0

One mole of non-ideal gas undergoes a change of state (1.0 atm, 3.0 L, 200 K) to (4.0 atm, 5.0 L, 250 K) with a change in internal energy ΔU = 40 L-atm. The change in enthalpy of the process in L-atm is :

A
43

B
57

C
42

D
none of these

Solution

When both P and V are changing, \(\Delta H=\Delta U+\Delta(P V)=\Delta U+\left(P_{2} V_{2}-P_{1} V_{1}\right)=40+(20-3)=57 L a t m\)
Question 8
1 / -0

A sample of oxygen gas expands its volume from 3 L to 5 L against a constant pressure of 3 atm. If work done during expansion is used to heat 10 mole of water initially present at 290 K, its final temperature will be (specific heat capacity of water = 4.18 J/kg):

A
292.0 k

B
298.0 k

C
290.8 k

D
293.7 k

Solution

Work done during expansion is equal to the heat supplied to water to raise the temperature. \(P\left(V_{2}-V_{1}\right)=m S \Delta T\)
\(3 \times 2 \times 101.3=180 \times 4.18\left(T_{f}-290\right)\)
\(T_{f}=290.8 K\)
Question 9
1 / -0

A system absorbs 300 cal of heat, its volume doubles and temperature rises from 273 to 298 k, the work done on the surrounding is 200 cal. ΔE for the above reaction is :

A
100 cal

B
500 cal

C
-500 cal

D
-100 cal

Solution

As per the sign convention, the heat absorbed by the system has positive sign. Hence, \(q=+300 \mathrm{cal}\)
When work is done by the system on the surroundings, it has negative sign. Thus \(w=-200\) cal. The expression for the change in the internal energy is \(\Delta E=q+w=+300 \mathrm{cal}-200 \mathrm{cal}=100 \mathrm{cal}\)
Thus, the net internal energy of the system increases by 100 cal.
Question 10
1 / -0

Certain amount of a gas confined in a piston-filled cylinder is heated from 27⁰C to 127⁰C keeping its pressure constant and the gas expanded against a constant pressure doing 4.157 kJ of work on surroundings. Find the number of moles of gas present in the cylinder ?

A
4

B
5

C
6

D
7

Solution

\(W=-P \Delta V\)
\(W=-n R \Delta T\)
or, \(n=\frac{4.157 \times 10^{3}}{8.314 \times 100}=5\)

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Re-Attempt Weekly Quiz Competition

Chemistry Test - 46

Result

Chemistry Test - 46

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SHARING IS CARING

Question 1

Lowering of energy in the process is the only criterion for spontaneity.

For a spontaneous process in an isolated system, the change in entropy is positive.

Endothermic process are never spontaneous.

Exothermic process are always spontaneous.

Solution

Question 2

2R

3R/2

5R/2

0

Solution

Question 3

Br2(g)

Cl2(g)

H2O(g)

CH4(g)

Solution

Question 4

4.98 kJ

11.47 kJ

-11.47 kJ

0kJ

Solution

Question 5

0.2 kJ/mol

2.0 kJ/mol

1.0 kJ/mol

10 kJ/mol

Solution

Question 6

Lowering of energy in the process is the only criterion for spontaneity.

For a spontaneous process in an isolated system, the change in entropy is positive.

Endothermic process are never spontaneous.

Exothermic process are always spontaneous.

Solution

Question 7

43

57

42

none of these

Solution

Question 8

292.0 k

298.0 k

290.8 k

293.7 k

Solution

Question 9

100 cal

500 cal

-500 cal

-100 cal

Solution

Question 10

4

5

6

7

Solution

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Br₂(g)

Cl₂(g)

H₂O(g)

CH₄(g)