Kinetic Theory Test

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Kinetic Theory Test - 48

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Question 1
1 / -0

A diatomic gas is heated at certain pressure. What fraction of the heat energy is used to increase the internal energy?

A
$$3/5$$

B
$$3/7$$

C
$$5/7$$

D
$$5/9$$

Solution

$$f=\displaystyle\frac{\Delta V}{(\Delta Q_p)}=\frac{\Delta Q_v}{(\Delta Q_p)_p}=\frac{\mu Cv\Delta T}{\mu Cp\Delta T}=\frac{C_v}{C_p}$$
Now $$C_p/C_v$$ is 7/5 for a diatomic gas.
Thus the fraction of heat energy used to increase the internal energy is given by $$5/7$$
Question 2
1 / -0

If the average kinetic energy of a molecule of a hydrogen gas at $$300$$K is E, then the average kinetic energy of a molecule of a nitrogen gas at the same temperature is:

A
$$7$$E

B
$$E/14$$

C
$$14$$E

D
$$E/7$$

E
E

Solution

We know that,
Kinetic energy $$\propto$$ $$ T$$
In the question, temperature is same in both conditions. So, the kinetic energy in the second condition is E.
Question 3
1 / -0

The gases are at absolute temperature $$300^oK$$ and $$350^oK$$ respectively. The ratio of average kinetic energy of their molecules is

A
$$6:7$$

B
$$36:49$$

C
$$7:6$$

D
$$343: 216$$

Solution

From the relation,
$$\overline{KE}=\displaystyle\frac{3}{2}kT$$
$$\Rightarrow \displaystyle\frac{(\overline{KE})_1}{(\overline{KE})_2}=\frac{T_1}{T_2}$$ (become k is constant)
Given, $$T_1=300K, T_2=350$$K
$$\therefore \displaystyle\frac{({KE})_1}{(KE)_2}=\frac{300}{350}=\frac{6}{7}$$.

Question 4

1 / -0

Match the List - I with List - II:

List-I	List-II
(I) Translation kinetic energy	1. $$\frac{3}{2}P$$
(II) Rotational kinetic energy of $$CO_2$$	2. 15/13
(III) Translation kinetic energy per unit volume	3. 7/5
(IV) $$\lambda$$ for $$CO_2$$ at very high temperature	4. Function of T only 5. RT 6. $$\frac{3}{2} RT$$

(I) - 3 (II) - 4 (III) - 5 (IV) - 1

(I) - 4 (II) - 5 (III) - 1 (IV) - 2

(I) - 5 (II) - 6 (III) - 2 (IV) - 3

(I) - 6 (II) - 1 (III) - 3 (IV) - 4

Solution

(I) Translation kinetic energy 7/5

(II) Rotational kinetic energy of $$CO_2$$ Function of T only 2. 15/13

(III) Translation kinetic energy per unit volume RT

(IV) $$\lambda$$ for $$CO_2$$ at very high temperature $$\frac{3}{2}P$$

Hence option A is correct.

Question 5
1 / -0

A container is filled with $$20$$ moles of an ideal diatomic gas at absolute temperature $$T$$. When heat is supplied to gas, temperature remains constant but $$8$$ moles dissociate into atoms. Heat energy given to gas is

A
$$4RT$$

B
$$6RT$$

C
$$3RT$$

D
$$5RT$$

Solution

Average Energy of 20 moles of diatomic gas $$= n \times \cfrac{5}{2} \times RT$$$$= 20 \times \cfrac{5}{2} \times RT = 50RT$$
Average Energy of $$12$$ $$moles$$ of diatomic gas $$+ 16$$ $$moles$$ of monoatomic gas $$(8$$ $$moles \rightleftharpoons 16$$ $$moles)$$
$$= 12 \times \cfrac{5}{2} \times RT$$$$ + 16 \times \cfrac{3}{2} \times RT = 54RT$$
Heat supplied to the gas $$= 54 RT - 50 RT = 4 RT$$
Question 6
1 / -0

A sample of gas in a box is at pressure $${ P }_{ 0 }$$ and temperature $${ T }_{ 0 }$$. If number of molecules is doubled and total kinetic energy of the gas kept constant then final temperature and pressure will be

A
$$T_{ 0 }\ .\ P_{ 0 }$$

B
$$T_{ 0 }\ .\ 2P_{ 0 }$$

C
$$\dfrac { { T }_{ 0 } }{ 2 } .{ 2P }_{ 0 }$$

D
$$\dfrac { { T }_{ 0 } }{ 2 } .{ P }_{ 0 }$$

Solution

Total Kinetic Energy = $$\cfrac{3}{2}nRT$$

$$\therefore$$ If kinetic energy is constant;

$$\cfrac{3}{2}n_1RT_1$$ $$=\cfrac{3}{2}n_2RT_2$$

If $$n_2 = 2n_1$$

$$T_2 = \cfrac{T_1}{2}$$ ($$ \cfrac{T_0}{2}$$)

Again, $$\cfrac{P_1V_1}{T_1} = $$$$\cfrac{P_2V_2}{ T_2}$$

$$P_2 = P_1 $$ $$(P_0)$$

$$\cfrac{T_0}{2},$$$$(P_0)$$ are the final temperature and pressure.
Question 7
1 / -0

The mean kinetic energy of one mole of gas per degree of freedom (on the basis of kinetic theory of gases) is

A
$$\cfrac { 1 }{ 2 } kT$$

B
$$\cfrac { 3 }{ 2 } kT\quad $$

C
$$\cfrac { 3 }{ 2 } RT\quad $$

D
$$\cfrac { 1 }{ 2 } RT$$

Solution

On the bases of kinetic energy of gases, the mean kinetic energy of one mole of gas per degree of freedom is 12RT
where R is the universal gas constant.
On the basis kinetic energy of a molecule of gas per degree of freedom is 12kT
where k is the Boltzmann constant
Question 8
1 / -0

The specific heat of a gas is found to be $$0.075$$ calories at constant volume and its formula $$wt$$ is $$40$$. The atomicity of the gas would be:

A
one

B
two

C
three

D
four

Solution

Specific heat of gas is defined as the amount of heat required to raise the temperature of 1 g of substance by 1 K.

Given: $$C_V=0.075\ cal/g/K$$. Therefore $$0.075\ cal$$ is absorbed by 1 g of the gas at constant volume.

Formula wt of gas is 40 g. So total heat absorbed by the gas is:

$$C_V=40\times 0.075=3\ cal/K/mol$$.

Since , $$C_p-C_V=R$$ and $$R=1.987\ cal/K/mol$$

$$\therefore\ C_p=C_V+R=3+1.987\approx 5$$

Calculating $$C_p/C_V=5/3=1.66$$

Hence the gas is monatomic. Therefore option A is correct.
Question 9
1 / -0

The temperature (T) of one mole of an ideal gas varies with its volume (V) as $$T = -\alpha { V }^{ 3 }+\ \beta { V }^{ 2 }$$, where $$\alpha$$ and $$\beta$$ are positive constants. The maximum pressure of gas during this process is

A
$$\dfrac { \alpha \beta }{ 2R } $$

B
$$\dfrac { { \beta }^{ 2 }R }{ 4\alpha }$$

C
$$\dfrac { (\alpha +\beta )R }{ { 2\beta }^{ 2 } }$$

D
$$\dfrac { { \alpha }^{ 2 }R }{ { 2\beta } }$$

Solution

$$T=-\alpha V^3+\beta V^2\\ \cfrac{PV}{R}=-\alpha V^2+\beta V^2\\ P=RV(-\alpha V+\beta)$$

For maximum pressure

$$\cfrac{dP}{dV}=0\\ \Rightarrow -2\alpha V+\beta =0\\ \Rightarrow V=\cfrac{\beta }{2\alpha}\\ P=R[-\alpha(\cfrac{\beta }{2\alpha})^2+\cfrac{\beta }{2\alpha}\times \beta]\\P=R[\cfrac{-\beta^2}{4\alpha}+\cfrac{\beta^2}{2\alpha}]\\P=\cfrac{R\beta^2}{4\alpha}$$
Question 10
1 / -0

X g of ice at $$0^o$$C is added to $$340$$g of water at $$20^o$$C. The final temperature of the resultant mixture is $$5^o$$C. The value of X (in g) is closest to. [Heat of fusion of ice$$=333\ J/g$$; Specific heat of water $$=4.184\ J/(g. K)$$].

A
$$80.4$$

B
$$52.8$$

C
$$120.6$$

D
$$60.3$$

Solution

Keeping in mind that heat loss by warmer must be equal to the heat gained by colder. We have this :

Heat to melt ice (heat of fusion) $$+$$ Heat to warm water from $$0^oC$$ to $$5^oC$$ $$=$$ Heat lost by warm water

$$333\times X+X\times (5-0)\times (4.184)=340\times (4.184)\times (20-5)$$

$$X=60.29\ g \approx 60.3\ g$$

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Kinetic Theory Test - 48

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Kinetic Theory Test - 48

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SHARING IS CARING

Question 1

$$3/5$$

$$3/7$$

$$5/7$$

$$5/9$$

Solution

Question 2

$$7$$E

$$E/14$$

$$14$$E

$$E/7$$

E

Solution

Question 3

$$6:7$$

$$36:49$$

$$7:6$$

$$343: 216$$

Solution

Question 4

(I) - 3 (II) - 4 (III) - 5 (IV) - 1

(I) - 4 (II) - 5 (III) - 1 (IV) - 2

(I) - 5 (II) - 6 (III) - 2 (IV) - 3

(I) - 6 (II) - 1 (III) - 3 (IV) - 4

Solution

Question 5

$$4RT$$

$$6RT$$

$$3RT$$

$$5RT$$

Solution

Question 6

$$T_{ 0 }\ .\ P_{ 0 }$$

$$T_{ 0 }\ .\ 2P_{ 0 }$$

$$\dfrac { { T }_{ 0 } }{ 2 } .{ 2P }_{ 0 }$$

$$\dfrac { { T }_{ 0 } }{ 2 } .{ P }_{ 0 }$$

Solution

Question 7

$$\cfrac { 1 }{ 2 } kT$$

$$\cfrac { 3 }{ 2 } kT\quad $$

$$\cfrac { 3 }{ 2 } RT\quad $$

$$\cfrac { 1 }{ 2 } RT$$

Solution

Question 8

one

two

three

four

Solution

Question 9

$$\dfrac { \alpha \beta }{ 2R } $$

$$\dfrac { { \beta }^{ 2 }R }{ 4\alpha }$$

$$\dfrac { (\alpha +\beta )R }{ { 2\beta }^{ 2 } }$$

$$\dfrac { { \alpha }^{ 2 }R }{ { 2\beta } }$$

Solution

Question 10

$$80.4$$

$$52.8$$

$$120.6$$

$$60.3$$

Solution

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