Thermodynamics Test

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Thermodynamics Test - 77

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Question 1
1 / -0

$$N$$ moles of a monoatomic gas is carried round the reversible rectangular cycle $$ABCDA$$ as shown in the diagram. The temperature at $$A$$ is $${T}_{0}$$. The thermodynamic efficiency of the cycle is:

A
$$40$$%

B
$$50$$%

C
$$60$$%

D
$$70$$%

Solution

Heat absorbed $$=$$ Work done by gas at constant pressure
$$=2{P}_{0}(2{V}_{0}-{V}_{0})=2{P}_{0}{V}_{0}$$
Net work done by the gas $$=$$ Work done by the gas$$-$$ Work done on the gas
$$=2{P}_{0}{V}_{0}-{P}_{0}{V}_{0}$$ [Net work done in a cyclic process is area of the loop]
$$={P}_{0}{V}_{0}$$
So, efficiency $$=\cfrac{{P}_{0}{V}_{0}}{2{P}_{0}{V}_{0}}\times 100=50$$%
Question 2
1 / -0

The relation between internal energy U, pressure P and Volume V of a gas in an adiabatic process is $$U=a+bPV$$, where a and b are positive constants. What is the value of $$\gamma$$

A
$$\dfrac { a }{ b } $$

B
$$\dfrac { b+1 }{ b } $$

C
$$\dfrac { a+1 }{ a } $$

D
$$\dfrac { b }{ a } $$

Solution
Question 3
1 / -0

The relation between internal energy U, pressure P and volume V of a gas in an adiabatic process is U=a+bPV, where a and b are positive constants. What is the value of $$\gamma$$?

A
$$\dfrac{a}{b}$$

B
$$\dfrac{b+1}{b}$$

C
$$\dfrac{a+1}{a}$$

D
$$\dfrac{b}{a}$$

Solution

In an adiabatic process no heat is exchange. i.e. $$(dQ = 0)$$

$$dQ = dU + PdV$$

$$0 = dU + Pd V$$

$$d (a + bPV) + PdV = 0$$ ... ($$V = a + b PV$$..(given))

$$bPdV + bVdP + PdV = 0$$

$$(b + 1) PdV + bVd P = 0$$

$$(b + 1)(dV/V) + b (dP / P ) = 0$$ ... (dividing both by PV)

Integrating we get,

$$(b + 1) \log V + b(\log P) =$$ constant

$$\log V^{(b + 1)} + \log P^b = $$ constant

$$V^{(b + 1)} \times P^b = $$ constant

Multiply power by $$\dfrac{1}{b}$$

$$PV^ {((b + 1) /b)} = constant = PV^{\gamma}$$

$$\gamma = \dfrac{b + 1}{b}$$

Hence option b is correct.
Question 4
1 / -0

A monoatomic ideal gas at temperature $$T_{1}$$, is enclosed in a cylinder with a frictionless piston. The gas is allowed to expanded adiabatically to a temperature $$T_{2}$$ by releasing the piston suddenly. If $$L_{1}$$, and $$L_{2}$$ are lengths of the gas columns before and after expansion respectively, then $$\frac{T_{1}}{T_{2}}$$ is given by:

A
$$(\frac{L_{1}}{L_{2}})^{\frac{L}{3}}$$

B
$$(\frac{L_{1}}{L_{2}})$$

C
$$(\frac{L_{2}}{L_{1}})$$

D
$$(\frac{L_{2}}{L_{1}})^{\frac{2}{3}}$$

Solution

For adiabatic process $$Tv^{\gamma - 1}$$ =
Let cross section area be $$A$$
initial volume $$v_1 = L_1A$$
Final volume $$v_2 = L_2 A$$
$$T_1 V_1^{\gamma - 1} = T_2 V_2 ^{\gamma - 1}$$
$$\dfrac{T_1}{T_2} = \left(\dfrac{V_2}{V_1}\right)^{\gamma - 1}$$
$$\dfrac{T_1}{T_2} = \left(\dfrac{L_2 A}{L_1 A}\right)^{\gamma - 1}$$
$$\dfrac{T_1}{T_2} = \left(\dfrac{L_2}{L_1}\right)^{\gamma - 1} = \left(\dfrac{L_2}{L_1} \right)^{5/3 -1} = \left(\dfrac{L_2}{L_1} \right)^{2/3}$$
option $$D$$
Question 5
1 / -0

In a thermodynamic process pressure of a fixed mass of a gas is changed in such a manner that the gas releases $$20J$$ of heat and $$8J$$ of work is done on the gas. If initial internal energy of the gas was $$30J$$, what will be the fixed internal energy?

A
$$9J$$

B
$$27J$$

C
$$18J$$

D
$$12J$$

Solution

$$dQ=-20J\\ dW=-8J\\ { V }_{ i }=30J\\ dQ=dV+dW\\ dQ=({ V }_{ f }-{ V }_{ i })+dW\\ -20=({ V }_{ f }-30)-8\\ -12={ V }_{ f }-30\\ { V }_{ f }=18$$
$$\therefore$$ Final internal energy=$$18J$$
Question 6
1 / -0

A ring shaped tube contains two ideal gases with equal masses and molar masses $$M_1=32$$ and $$M_2=28$$. The gases are separates by one fixed partition and another movable stopper $$S$$ which can move freely without friction inside the ring. The angle $$\alpha$$ as shown in the figure is _________ degrees.

A
$$291$$

B
$$219$$

C
$$129$$

D
$$192$$

Solution

Pressure on both the sides will be equal
$$\Rightarrow P_1=P_2$$
$$\Rightarrow \dfrac{n_1RT}{V_1}=\dfrac{n_2RT}{V_2}$$
$$\Rightarrow \dfrac{n_1}{V_1}=\dfrac{n_2}{V_2}$$
We know, n=$$\dfrac{m}{M}$$
$$\Rightarrow \dfrac{m}{M_1V_1}=\dfrac{m}{M_2V_2}$$
$$\Rightarrow \dfrac{V_2}{V_1}=\dfrac{M_2}{M_1} =\dfrac{32}{28}=\dfrac{8}{7}$$
But, $$\dfrac{V_2}{V_1} = \dfrac{\alpha}{360°- \alpha}$$
Therefore,$$ \dfrac{8}{7}=\dfrac{\alpha}{360°- \alpha}$$
$$\Rightarrow 8(360°-\alpha))=7(\alpha)$$
$$\Rightarrow 15\alpha =360°×8$$
therefore, $$\alpha = 192$$
Hence, the correct option is D.
Question 7
1 / -0

Three samples of the same gas A, B and C$$\left( \gamma =3/2 \right) $$ have equal volume initially. Now, the volume of each sample is doubled. For A, the process is adiabatic; for B, it is isobaric and for C, the process is isothermal. If the final pressures are equal for all the three samples, the ratio of their initial pressures is

A
$$2:1:\sqrt { 2 } $$

B
$$2\sqrt { 2 } :1:2$$

C
$$\sqrt { 2 } :1:2$$

D
$$\sqrt { 2 } :2:1$$

Solution

Let the initial pressure of the three sample be $$P_A$$.
$$P_B $$ and $$P_C$$
then $$P_A(V)^{\frac{3}{2}} = (2V)^{\frac{3}{2}}P$$

$$P_B = P$$

$$ \, P_C (V) = P(2V)$$

$$\therefore P_A : P_B : P_C = (2)^{3/2},\, 1:2 = 2 \sqrt{2} : 1 : 2$$
Question 8
1 / -0

A fixed thermally conducting cylinder has a radius R and height $$L_0$$. The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is $$P_0$$.
While the piston is at a distance $$2L$$ from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is?

A
$$\left(\dfrac{2P_0\pi R^2}{\pi R^2P_0+Mg}\right)(2L)$$

B
$$\left(\dfrac{P_0\pi R^2-Mg}{\pi R^2P_0}\right)(2L)$$

C
$$\left(\dfrac{P_0\pi R^2+Mg}{\pi R^2P_0}\right)(2L)$$

D
$$\left(\dfrac{P_0\pi R^2}{\pi R^2P_0-Mg}\right)(2L)$$

Solution
Question 9
1 / -0

When an ideal monoatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases the internal energy of the gas is

A
$$2/5$$

B
$$3/5$$

C
$$3/7$$

D
$$5/7$$

Solution

By first law of thermodynamics
$$Q = U + W$$
$$= \dfrac {f}{2} nRT + P\int dV = \dfrac {f}{2} nRT + PV$$
or $$Q = \dfrac {f}{2} nRT + nRT (\because PV = nRT)$$
$$= \dfrac {3}{2}nRT + nRT = \dfrac {5}{2}nRT$$
$$\therefore$$ Fraction of heat energy supplied $$= \dfrac {U}{Q}$$
$$= \dfrac {(3/2)nRT}{(5/2)nRT} = \dfrac {3}{5}$$.
Question 10
1 / -0

The characterisic distance at which quantum gravitational effects are significant, the Planck length, can be determined from suitable combination of the fundamental physical constants $$G$$. $$h$$ and $$c$$. Which of the following correctly gives the Planck length?

A
$$G{h}^{2}{c}^{3}$$

B
$${G}^{2}hc$$

C
$${G}^{1/2}{h}^{2}c$$

D
$${ \left( \cfrac { Gh }{ { c }^{ 3 } } \right) }^{ 1/2 }$$

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Re-Attempt Weekly Quiz Competition

Thermodynamics Test - 77

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Thermodynamics Test - 77

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SHARING IS CARING

Question 1

$$40$$%

$$50$$%

$$60$$%

$$70$$%

Solution

Question 2

$$\dfrac { a }{ b } $$

$$\dfrac { b+1 }{ b } $$

$$\dfrac { a+1 }{ a } $$

$$\dfrac { b }{ a } $$

Solution

Question 3

$$\dfrac{a}{b}$$

$$\dfrac{b+1}{b}$$

$$\dfrac{a+1}{a}$$

$$\dfrac{b}{a}$$

Solution

Question 4

$$(\frac{L_{1}}{L_{2}})^{\frac{L}{3}}$$

$$(\frac{L_{1}}{L_{2}})$$

$$(\frac{L_{2}}{L_{1}})$$

$$(\frac{L_{2}}{L_{1}})^{\frac{2}{3}}$$

Solution

Question 5

$$9J$$

$$27J$$

$$18J$$

$$12J$$

Solution

Question 6

$$291$$

$$219$$

$$129$$

$$192$$

Solution

Question 7

$$2:1:\sqrt { 2 } $$

$$2\sqrt { 2 } :1:2$$

$$\sqrt { 2 } :1:2$$

$$\sqrt { 2 } :2:1$$

Solution

Question 8

$$\left(\dfrac{2P_0\pi R^2}{\pi R^2P_0+Mg}\right)(2L)$$

$$\left(\dfrac{P_0\pi R^2-Mg}{\pi R^2P_0}\right)(2L)$$

$$\left(\dfrac{P_0\pi R^2+Mg}{\pi R^2P_0}\right)(2L)$$

$$\left(\dfrac{P_0\pi R^2}{\pi R^2P_0-Mg}\right)(2L)$$

Solution

Question 9

$$2/5$$

$$3/5$$

$$3/7$$

$$5/7$$

Solution

Question 10

$$G{h}^{2}{c}^{3}$$

$${G}^{2}hc$$

$${G}^{1/2}{h}^{2}c$$

$${ \left( \cfrac { Gh }{ { c }^{ 3 } } \right) }^{ 1/2 }$$

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