Thermodynamics Test

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Thermodynamics Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

A certain mass of gas is taken from an initial thermodynamic state A to another state B by process I and II. In process I the gas does 5 joules of work and absorbs 4 joules of heat energy. In process II, the gas absorbs 5 joules of heat. The work done by the gas in process II (see figure) is

A
+6 joules

B
-6 joules

C
+4 joules

D
-4 joules

Solution

Internal energy doesnt depend on path.
$$\therefore Q=U+W$$
in path $$I$$
$$Q=u, w=+5$$
$$U=-1\ J$$
in path $$II$$
$$Q=5, U=-1$$
$$\therefore W=5-(-1)=6\ J$$
Question 2
1 / -0

Which of the following parameters does not charaterize the thermodynamic state of matter?

A
temperature

B
pressure

C
work

D
volume

Solution

From the above options Work depends on the shift from one state to another
Question 3
1 / -0

$$5.6\ litre$$ of helium gas at STP is adiabatically compressed to $$0.7\ litre$$. Taking the initial temperature to be $$T_{1}$$, the work done in the process is

A
$$\dfrac {9}{8}RT_{1}$$

B
$$\dfrac {3}{2}RT_{1}$$

C
$$\dfrac {15}{8}RT_{1}$$

D
$$\dfrac {9}{2}RT_{1}$$

Solution

Number of moles of $$He = 5.6/ 22.4 = 1/4$$
Now $$T(5.6)^{\gamma - 1} = T_{2} (0.7)^{\gamma - 1}$$
$$T_{1} = T_{2} \left (\dfrac {1}{8}\right )^{2/3} \Rightarrow 4T_{1} = T_{2}$$
Work done $$= \dfrac {nR[T_{2} - T_{1}]}{\gamma - 1} = \dfrac {\dfrac {1}{4} R[3T_{1}]}{\dfrac {2}{3}} = \dfrac {9}{8} RT_{1}$$.
Question 4
1 / -0

"Heat cannot by itself flow from a body at lower temperature to a body at higher temperature" is a statement of the consequence of

A
second law of thermodynamics

B
conservation of momentum

C
conservation of mass

D
first law of thermodynamics

Solution

Hint: The second law of thermodynamics dictates that an isolated system’s entropy will not decrease with time. And first law states that energy cannot be created not be destroyed and that it is simply transferred from one form to the other

Step 1: Explanation:

The Clausius statement on Second law of thermodynamics states that, “Heat cannot flow from a cold body to a hot body without the performance of work by some external agency.” This is in direct correlation to the second law.

$$\textbf{Hence, the correct option is (A)}$$
Question 5
1 / -0

Which of the following complexes has the least molar conductivity in the solution?

A
$$CO{ CI }_{ 3 }.3{ NH }_{ 3 }$$

B
$$CO{ CI }_{ 3 }.4{ NH }_{ 3 }$$

C
$$CO{ CI }_{ 3 }.5{ NH }_{ 3 }$$

D
$$CO{ CI }_{ 3 }.6{ NH }_{ 3 }$$

Solution

$$COCl_3 . 3NH_2\to $$ It give three ammonia in water.
Question 6
1 / -0

An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity $$C$$ remains constant. If during this process the relation of pressure $$P$$ and volume $$V$$ is given by $${PV}^{n}=$$ constant, then $$n$$ is given by (Here $${C}_{P}$$ ad $${C}_{V}$$ are molar specific heat at constant pressure and constant volume, respectively):

A
$$n=\cfrac { C-{ C }_{ P } }{ C-{ C }_{ V } } $$

B
$$n=\cfrac { { C }_{ P }-C }{ C-{ C }_{ V } } $$

C
$$n=\cfrac { C-{ C }_{ V } }{ C-{ C }_{ P } } $$

D
$$n=\cfrac { { C }_{ P } }{ { C }_{ V } } $$

Solution

The correct option is A.

Given,

$$pressure= p, Volume=V$$

So, $$C=c_v+\dfrac{R}{1-n}\Rightarrow 1-n=\dfrac{R}{C-C_v}\Rightarrow n=1-\dfrac{R}{C-C_v}=\dfrac{C-(C_v+R)}{C-C_v}$$

$$=\dfrac{C-C_p}{C-C_v}$$
Question 7
1 / -0

What is the work done by $$0.2$$ mole of a gas at room temperature to double volume during isobaric process (take $$R=2 cal \ mol^{-1} K^{-1}$$)

A
$$30$$ cal

B
$$40$$ cal

C
$$120$$ cal

D
$$160$$ cal

Solution

Work done $$W = P(V_f-V_i) = P(2V-V) = PV$$
So, we know that the work done $$W = PV =nRT$$
$$W=0.2\times2\times300=120\ cal$$
Where $$n$$ is the mole of gas, $$T$$ is the temprature,
Question 8
1 / -0

One end of a $$0.25\ m$$ long metal bar is in steam and the other end is in contact with ice. If $$12\ g$$ of ice melts per minute, what is the thermal conductivity of the metal? Given cross-section of the bar $$= 5\times 10^{-4} m^{2}$$ and latent heat of ice is $$80\ cal/g$$.

A
$$80\ cal/s-m-^{\circ}C$$

B
$$90\ cal/s-m-^{\circ}C$$

C
$$70\ cal/s-m-^{\circ}C$$

D
$$60\ cal/s-m-^{\circ}C$$

Solution
Question 9
1 / -0

Which heat is produced throughout the conducting wire?

A
Petlier heat

B
Thomson effect heat

C
Joule heat

D
none of these

Solution

As the heat produced in the conductor by following Joule's law of heating or cooling. Sp, the produced heat is Joule heat.
Question 10
1 / -0

Magnitude of Seebeck emf between the junctions does not depend on

A
thermocouple

B
temperature of cold junction

C
temperature of hot function

D
neutral temperature

Solution

The Seeback effect is the direct conversion of heat into electricity in which potential difference is needed to calculate the current density so the potential of two junctions must be known. It has nothing to do with neutral temperature.

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Re-Attempt Weekly Quiz Competition

Thermodynamics Test - 53

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Thermodynamics Test - 53

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SHARING IS CARING

Question 1

+6 joules

-6 joules

+4 joules

-4 joules

Solution

Question 2

temperature

pressure

work

volume

Solution

Question 3

$$\dfrac {9}{8}RT_{1}$$

$$\dfrac {3}{2}RT_{1}$$

$$\dfrac {15}{8}RT_{1}$$

$$\dfrac {9}{2}RT_{1}$$

Solution

Question 4

second law of thermodynamics

conservation of momentum

conservation of mass

first law of thermodynamics

Solution

Question 5

$$CO{ CI }_{ 3 }.3{ NH }_{ 3 }$$

$$CO{ CI }_{ 3 }.4{ NH }_{ 3 }$$

$$CO{ CI }_{ 3 }.5{ NH }_{ 3 }$$

$$CO{ CI }_{ 3 }.6{ NH }_{ 3 }$$

Solution

Question 6

$$n=\cfrac { C-{ C }_{ P } }{ C-{ C }_{ V } } $$

$$n=\cfrac { { C }_{ P }-C }{ C-{ C }_{ V } } $$

$$n=\cfrac { C-{ C }_{ V } }{ C-{ C }_{ P } } $$

$$n=\cfrac { { C }_{ P } }{ { C }_{ V } } $$

Solution

Question 7

$$30$$ cal

$$40$$ cal

$$120$$ cal

$$160$$ cal

Solution

Question 8

$$80\ cal/s-m-^{\circ}C$$

$$90\ cal/s-m-^{\circ}C$$

$$70\ cal/s-m-^{\circ}C$$

$$60\ cal/s-m-^{\circ}C$$

Solution

Question 9

Petlier heat

Thomson effect heat

Joule heat

none of these

Solution

Question 10

thermocouple

temperature of cold junction

temperature of hot function

neutral temperature

Solution

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