Thermodynamics Test

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Thermodynamics Test - 43

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Question 1
1 / -0

In a thermodynamic process pressure of a fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat and 8 J of work is done on the gas if the initial internal energy of the gas was 30 J. The final internal energy wilI be:

A
26 J

B
15 J

C
18 J

D
9 J

Solution

dQ = -20J, dW = -8 J, $$U_1 = 30J$$
By 1st law of thermodynamics,
dQ = dU + dW
$$\Rightarrow $$ dU = dQ-dW
= -20 + 8 = -12J
Final internal energy, $$U_2 = U_1 + dU$$
= 30-12 = 18 J
Question 2
1 / -0

Internal energy of a gas increases when

A
The change is cyclic

B
It absorbs heat

C
The change is adiabatic

D
None

Solution

An adiabatic process is one in which no heat is gained or lost by the system. The first law of thermodynamics with Q = 0 shows that all the change in internal energy is in the form of work done. Internal energy of a gas increases when the change is adiabatic.
Question 3
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Zeroth law of thermodynamic signifies:

A
$$Mass$$

B
$$Heat$$

C
$$Temperature$$

D
$$Inertia$$

Solution

The "zeroth law" states that if two systems are at the same time in thermal equilibrium with a third system, they are in thermal equilibrium with each other.
So,Zeroth law of thermodynamic signifies Temperature.
hence, option C is correct.
Question 4
1 / -0

If heat is supplied to the system it :

A
Does work only

B
Increase the internal energy

C
Either does work or increase the internal energy or both

D
Docs nothing

Solution

The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. The law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed. The first law is often formulated.

$${\displaystyle \Delta U=Q-W.} {\displaystyle \Delta U=Q-W.}$$
It states that the change in the internal energy ΔU of a closed system is equal to the amount of heat Q supplied to the system, minus the amount of work W done by the system on its surroundings. An equivalent statement is that perpetual motion machines of the first kind are impossible.
heat is supplied to the system it ither does work or increase the internal energy or both
Question 5
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Hydrogen and helium each having volume $$V$$ at same temperature $$T$$ and same pressure $$P$$ are mixed to have same volume $$V$$. The resulting pressure of the mixture will be:

A
$$R/2$$

B
$$P$$

C
$$2P$$

D
$$depending\ on\ the\ relative\ mass\ of\ the\ gases$$

Solution

According to boyle’s law, at constant temperature

$$ V\propto \frac{1}{P} $$

$$ $$ $${{P}_{1\,\,}}\,{{V}_{1\,}}\,=\,\,{{P}_{2\,\,}}{{V}_{2}}$$ = constant

Where V and P are volume and pressure of given gas

$$ {{P}_{1\,\,\,}}=\,\,P,\,\,{{V}_{1}}\,=\,V\,+\,V=\,2V $$

$$ {{V}_{2\,\,\,}}=V\,,\,{{P}_{2}}\,=\,? $$

$$ P\times \,\,2V\,=\,V\,\times \,{{P}_{2}} $$

$$ {{P}_{2}}\,=\,2\,\,P $$

_{hence, option C is correct.}
Question 6
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Two gases of equal masses are in thermal equilibrium. If $$P_a, P_b$$ and $$V_a, V_b$$ are their respective pressures and volumes, then which relation. is true?

A
$$2 P_aV_a = P_b V_b$$

B
$$2 P_a \neq V_a, P_b = V_b$$

C
$$\displaystyle \frac{P_a}{V_a} = \frac{P_b}{V_b}$$

D
$$P_a V_a = P_b V_b$$

Solution

Systems are in thermal equilibrium. So, it is an isothermal process.
So, $$P_aV_a=P_bV_b$$
Question 7
1 / -0

According to Boyle's law, PV = constant, at constant temperature. Which of the following graphs is correct?

A

B

C

D

Solution

The correct answer is option(D).
According to Boyle's law, PV = constant, this is satisfied in graph of option(D).
Question 8
1 / -0

At constant temperature for an ideal gas of a fixed mass, the variation of the product $$PV$$ with volume $$V$$ is graphically represented by the curve

A

B

C

D

Solution

Given :At constant temperature for an ideal gas of a fixed mass, the variation of the product PV with volume V is graphically represented by the curve.
$$T\rightarrow constant$$
Solution:
For an ideal gas we have relation as PV=nRT
Since temperature is constant and mass is also fixed then
PV will be constant irrespective the value of V
The Correct Opt=C
Question 9
1 / -0

Two bodies A and B are said to be in thermal equilibrium with each other, if

A
heat flows from A to B

B
heat flows from B to A

C
both the bodies lose equal amount of heat to the atmosphere

D
heat does not flow from either A or B

Solution

Explanation:
$$\bullet$$When thermal equilibrium is established, the temperature of the two bodies will be same. So, temperature of bodies A and B will be same.
$$\bullet$$At that time, no energy will be flow from A to B or from B to A. As there is no energy transferred, so heat will also not flow from both bodies A and B.
$$Answer:$$
Hence, option D is the correct answer.
Question 10
1 / -0

Figure shows pressure P-Y graph for a certain mass of a gas at two constant temperatures $$T_1$$ and $$T_2$$. Which of the inferences given below is correct?

A
$$T_1 = T_2$$

B
$$T_1 > T_2$$

C
$$T_1 < T_2$$

D
none of these

Solution

For a given volume V, pressure at curve $$T_1$$ will be lower than that at curve $$T_2$$. Therefore, $$P_2$$ > $$P_1$$ and so $$T_2$$ > $$T_1$$.