Thermodynamics Test

Result

Thermodynamics Test - 39

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Air is filled ina motor car tube at $$27^oC$$ temperature and $$2\ atm$$ pressure. If the tube suddenly bursts then the final temperature will be [given $$(1/2)^{2/7} = 0.82$$]

A
$$642 K$$

B
$$563 K$$

C
$$300 K$$

D
$$246 K$$

Solution

For an adiabatic process:
$$P^{1-\gamma}T^{\gamma} = constant$$
Initial pressure and temperatures are:
2atm and 300K respectively.
After the explosion the final pressure is 1atm
$$P_{1}^{1-\gamma}T_{1}^{\gamma} = P_{2}^{1-\gamma}T_{2}^{\gamma}$$
For air, majority of the gas is nitrogen which is diatomic
Hence $$\gamma = \dfrac{7}{2}$$

$$(2atm)^{1-(7/2)}(300)^{(7/2)} = (1\ atm)^{1-(7/2)}(T_{2}^{(7/2)})$$
i.e. We get $$T_{2} = 246K$$
Question 2
1 / -0

$$1\ g$$ mole of an ideal gas at $$STP$$ is subjected to a reversible adiabatic expansion to double its volume. Find the change in internal energy ($$\gamma = 1.4$$ )

A
$$11.695.5 J$$

B
$$769.5 J$$

C
$$1369.5 J$$

D
$$969.5 J$$

Solution

As it is an adiabatic process we use
$$T_1{ V }_{1 }^{ \gamma-1 }= T_2{V}_{2}^{\gamma - 1}$$
or $$T_2 =\displaystyle\dfrac {T_1 { V }_{1 }^{ \gamma-1 } }{ { V }_{2 }^{\gamma-1 } } = \dfrac{273}{(2)^{0.4}} = 207 K$$
Change the Internal Energy $$\displaystyle \Delta U= \dfrac{R}{(\gamma-1)}(T_1-T_2)$$
$$=\displaystyle \dfrac{8.31(273-207)}{1.4-1} = 1369.5 J$$
Question 3
1 / -0

Variation of molar specific heat of a metal with temperature is best depicted by

A

B

C

D

Solution

For $$T\rightarrow 0$$ =, that is, at low temperature, molar specific heat $$\propto$$ T$$^3$$ and high temperature it becomes constant = $$3 R$$. This is given by $$\text{Dulong and Petit law}$$.
Question 4
1 / -0

A mass of ideal gas at pressure $$P$$ is expanded isothermally to four times the original volume and then slowly compressed adiabatically to its original volume. Assuming $$\gamma$$ to be $$1.5$$, the new pressure of gas is

A
$$2P$$

B
$$P$$

C
$$4P$$

D
$$\displaystyle{\frac{P}{2}}$$

Solution

Let $$P$$ and $$V$$ be the initial pressure and volume of ideal gas. After isothermal expansion, pressure is $$P/4$$. So volume is $$4 V$$.
Let $$P_1$$ be the pressure after adiabatic compression then
$$P_1 V^{\gamma} = (P/4)(4V)^{\gamma}$$
$$P_1 = (\dfrac{P}{4}) (4)^{3/2} = 2P$$
Question 5
1 / -0

The pressure inside a tyre is $$4$$ times that of atmosphere. If the tyre burst suddenly at temperature $$300 K$$, what will be the new temperature?

A
$$300(4)^{7/2}$$

B
$$300(4)^{2/7}$$

C
$$300(2)^{7/2}$$

D
$$300(4)^{-2/7}$$

Solution

Under adiabatic change
$$\displaystyle{\frac{T_2}{T_1} = \left(\dfrac{P_1}{P_2}\right)^{\frac{1 - \gamma}{\gamma}}}$$
or $$\displaystyle{T_2 = T_1 (\dfrac{P_1}{P_2})^{(\frac{1- \gamma}{\gamma})}}$$
$$\therefore$$$$\displaystyle{T_2 = 300(4)^{\frac{1-(7/5)}{7/5}}}$$; $$\gamma = 1.4 = 7/5$$ for air
or $$T_2 = 300 (4)^{-2/7}$$
Question 6
1 / -0

At $$27^{o}C$$ a gas is compressed suddenly such that its pressure becomes ($$\dfrac{1}{8}$$)th of original pressure. Final temperature will be ($$\gamma$$ = 5/3)

A
$$450 K$$

B
$$300K$$

C
$$-142^{o}C$$

D
$$327^{o}$$

Solution

since gas is compressed suddenly, gas is undergone an adiabatic process
In adiabatic process,
$$P{ V }^{ \gamma }=constant$$
for an ideal gas,
$$\dfrac { PV }{ T } =constant$$
$$\therefore P{ (\dfrac { T }{ P } ) }^{ \gamma }={ P }^{ 1-\gamma }{ T }^{ \gamma }=constant$$
$$\therefore { { P }_{ 1 } }^{ 1-(5/3) }{ { T }_{ 1 } }^{ 5/3 }={ (\dfrac { { P }_{ 1 } }{ 8 } ) }^{ 1-(5/3) }{ T }^{ 5/3 }$$
$$\dfrac { 300 }{ T } ={ 4 }^{3/5 }$$
$$T=130.58K={ -142.58 }^{ \circ }C$$
Hence, option C is correct
Question 7
1 / -0

A diatomic gas initially at $$18^{o}C$$ is compressed adiabatically to one eightth of its original volume. The temperature after compression will be

A
$$18^{o}C$$

B
$$887^{o}C$$

C
$$327^{o}C$$

D
None of these

Solution

$$T_1 = 18^{o}C = (273 + 18) = 291\ K$$
and $$V_2 = V_1/8$$

We know that $$TV^{\gamma - 1}$$ = constant
or, $$T_2V_2^{\gamma - 1} = T_1V_1^{\gamma - 1}$$
$$\displaystyle{T_2 = T_1\left(\dfrac{V_1}{V_2}\right)^{\gamma - 1} = 291 \times (8)^{1.4 - 1}}= 668.5 K = 395.5^{o}C$$
Question 8
1 / -0

A perfect gas goes from state A to another state B by absorbing $$8 \times 10^5J$$ of heat and doing $$6.5 \times 10^5J$$ of external work. It is now transferred between the same two states in another process in which it absorbs $$10^5J$$ of heat. In the second process

A
work done by the gas is $$10^5J$$

B
work done on gas is $$10^5J$$

C
work done by gas is $$0.5 \times 10^5J$$

D
work done on the gas is $$0.5 \times 10^5J$$

Solution

$$dU = dQ - dW = (8 \times 10^{5} - 6.5 \times 10^{5} ) = 1.5 \times 10^{5}J$$
$$dW = dQ - dU = 10^{5} - 1.5\times 10^{5} = - 0.5 \times 10^{5}J$$
- ve sign indicates that work done on the gas is $$0.5 \times 10^5J$$
Question 9
1 / -0

$$80\ g$$ of water at $$30^o C$$ is poured on large block of ice at $$0^{\circ}$$C. The mass of ice that melts is
Given, Latent heat = $$334\ J/g$$
Specific heat of water $$= 4.2 Jg^{-1}C^{o-1}$$

A
$$1600\ g$$

B
$$30\ g$$

C
$$150\ g$$

D
$$80\ g$$

Solution

$$\Delta T=30^oC$$
Heat released by water is abosrbed by ice ,the final temperature of water will be $$0^oC$$
$$\Rightarrow 8\times1\times30=m\times80$$
$$\Rightarrow m=30g$$
Question 10
1 / -0

The volume of a gas is reduced adiabatically to $$\dfrac{1}{4}$$ of its volume at $$27^{o}C$$. If $$\gamma= 1.4$$ the new temperature is :

A
$$(300) 2^{0.4}K$$

B
$$(300) 2^{1.4}K$$

C
$$300 (4)^{0.4}K$$

D
$$300 (2)^{0.4}K$$

Solution

For an adiabatic change,
$$TV^{\gamma -1} =constant$$.....................(1)
where $$\gamma $$ is ratio of specific heats of gas.
From (1),
$$T_1V_1^{\gamma -1} =T_2V_2^{\gamma -1}$$
Here , $$T_1= 300K$$ , $$V_1=V$$ , $$V_2 =\dfrac{V}{4}$$
So, $$T_2 =( \dfrac{V_1}{V_2})^{\gamma -1}*1$$
putting the values , we get
$$T_2 =300(4)^{0.4} K$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Thermodynamics Test - 39

Result

Thermodynamics Test - 39

Score

-

Rank

-

SHARING IS CARING

Question 1

$$642 K$$

$$563 K$$

$$300 K$$

$$246 K$$

Solution

Question 2

$$11.695.5 J$$

$$769.5 J$$

$$1369.5 J$$

$$969.5 J$$

Solution

Question 3

Solution

Question 4

$$2P$$

$$P$$

$$4P$$

$$\displaystyle{\frac{P}{2}}$$

Solution

Question 5

$$300(4)^{7/2}$$

$$300(4)^{2/7}$$

$$300(2)^{7/2}$$

$$300(4)^{-2/7}$$

Solution

Question 6

$$450 K$$

$$300K$$

$$-142^{o}C$$

$$327^{o}$$

Solution

Question 7

$$18^{o}C$$

$$887^{o}C$$

$$327^{o}C$$

None of these

Solution

Question 8

work done by the gas is $$10^5J$$

work done on gas is $$10^5J$$

work done by gas is $$0.5 \times 10^5J$$

work done on the gas is $$0.5 \times 10^5J$$

Solution

Question 9

$$1600\ g$$

$$30\ g$$

$$150\ g$$

$$80\ g$$

Solution

Question 10

$$(300) 2^{0.4}K$$

$$(300) 2^{1.4}K$$

$$300 (4)^{0.4}K$$

$$300 (2)^{0.4}K$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.