Thermodynamics Test

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Thermodynamics Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

Two moles of Helium gas are taken over the cycle ABCDA as shown in the P-T diagram. Assuming the gas to be ideal the work done on the gas in taking it from $$\mathrm{A}$$ to $$\mathrm{B}$$ is :

A
$$200$$ $$\mathrm{R}$$

B
$$300$$ $$\mathrm{R}$$

C
$$400$$ $$\mathrm{R}$$

D
$$500$$ $$\mathrm{R}$$

Solution

Work done in isobaric process,$$\mathrm{W}_{\mathrm{A}\mathrm{B}}=P\Delta V=\mathrm{n}\mathrm{R}(\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}) =2\times \mathrm{R}(500-300)=400\mathrm{R}$$
Question 2
1 / -0

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

A
temperature

B
pressure

C
work

D
volume

Solution
Question 3
1 / -0

During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio $$\displaystyle \dfrac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$$ for the gas is:

A
$$ \dfrac{4}{3}$$

B
$$2$$

C
$$\dfrac{5}{3}$$

D
$$\dfrac{3}{2}$$

Solution
Question 4
1 / -0

An engine operates by taking n moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is :
(Take $$C_v$$ = 1.5 R, where R is gas constant )

A
0.32

B
0.24

C
0.15

D
0.08

Solution

$$\textbf{Solution}:$$
As we know the formula of work done which is equal to the
$$W = P_0V_0$$
Heat given will be equal to, = $${Q_{AB}} + {Q_{BC}}$$
Now by using the heat given formula mentioned above, we can write it as
$$\Rightarrow n{c_v}d{t_{AB}} + n{c_p}d{t_{BC}}$$
As we already know the value of $${C_v} = 1.5R$$
Therefore substituting this value in the heat given equation, we get,
$$= \dfrac{3}{2}\left( {nR{T_B} - nR{T_A}} \right) + \dfrac{5}{2}\left( {nR{T_C} - nR{T_B}} \right)$$
For the monatomic gas, the value will be 3
Now putting the values we had calculated above, we get
$$\Rightarrow \dfrac{3}{2}\left( {2{P_0}{V_0} - {P_0}{V_0}} \right) + \dfrac{5}{2}\left( {4{P_0}{V_0} - 2{P_0}{V_0}} \right)$$
On simplifying the solution, we get
$$\Rightarrow \dfrac{{13}}{2}{P_0}{V_0}$$
As we know the formula for efficiency can be given by
$$\eta = \dfrac{W}{{heat{\text{ gain}}}}$$
It can also be written in the following way and also substituting the values, we get
$$\Rightarrow \dfrac{{{P_0}{V_0}}}{{\dfrac{{\dfrac{{13}}{2}}}{{{P_0}{V_0}}}}}$$
On solving,
$$\Rightarrow \dfrac{2}{{13}} \Rightarrow \eta = 0.15$$

$$\textbf{Hence C is the correct option}$$
Question 5
1 / -0

The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is:

A
$$\dfrac{2}{3}$$

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

C
$$\dfrac{3}{5}$$

D
$$\dfrac{2}{5}$$

Solution

In isobaric process, heat supplied is $$Q=n{ C }_{ P }\Delta T$$.

Work done is $$W=P\Delta V=nR\Delta T$$.

Hence ratio is $$\dfrac { R }{ { C }_{ P } } =\dfrac { R }{ \dfrac { \gamma R }{ \gamma -1 } } =\dfrac { 2 }{ 5 } $$ [$$\because \gamma=\dfrac{5}{3}$$, for a monoatomic gas]
Question 6
1 / -0

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes adiabatic expansion, the average time of collision between molecules increases as $$V^q$$, where V is the volume of the gas. The value of q is :
$$\displaystyle \left( \gamma = \dfrac{C_p}{C_v} \right)$$

A
$$\displaystyle \dfrac{3 \gamma + 5}{6}$$

B
$$\displaystyle \dfrac{3 \gamma - 5}{6}$$

C
$$\displaystyle \dfrac{\gamma + 1}{2}$$

D
$$\displaystyle \dfrac{\gamma - 1}{2}$$

Solution

Fro an adiabatic process $$Tv^{r-1}=$$constant, we know that average time of collision between molecules
$$\tau=\dfrac{1}{n \pi \sqrt 2 V_{rms} d^2}$$
where $$n=$$ no, of molecules per unit volume
$$V_{rms}=rms$$ velocity of molecules
As $$n \propto \dfrac{1}{V}$$ and $$V_{rms} \propto \sqrt t$$
$$\tau \propto \dfrac{V}{\sqrt T}$$
Thus we can write:
$$n=k_1v^{-1}$$ and $$V_{rms}=K_2 T^{1/2}$$
where $$k_1$$ and $$k-2$$ are constant
for adiabatic process $$TV^{r-1}=$$ constant
Thus we can write
$$\tau \propto VT^{-1/2} \propto V(v^{1-r})^{-1/2}$$
or $$\tau \propto V^{\dfrac{r+1}{2}}$$
Average time between collision $$=\dfrac{means\ free\ path}{V_{rms}}$$
$$t=\dfrac{1}{\dfrac{\pi d^2N/V}{\sqrt{\dfrac{3RT}{M}}}}; t=\dfrac{CV}{\sqrt T} $$ where $$C=\dfrac{\sqrt M}{\pi d^2 B \sqrt{3R}}$$
$$\Rightarrow T \propto \dfrac{V^2}{t^2}$$
For adiabatic
$$TV^{\gamma-1}=k$$
$$\dfrac{V^2}{t^2}V^{\gamma -1}=k$$
$$\dfrac{V^{\gamma -1}}{t^2}=k,\ t \propto \dfrac{\gamma +1}{2}$$
so $$q=\dfrac{\gamma +1}{2}$$
Question 7
1 / -0

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

A
second law of thermodynamics

B
conservation of momentum

C
conservation of mass

D
first law of thermodynamics.

Solution

From Kelvin Plank statement and Claussius's statement we can say that heat can not be flow form lower temperature to higher temperature until and unless we are adding any external source like heat pump, which justifies $${SECOND \ LAW \ OF \ THERMODYNAMICS}$$
Question 8
1 / -0

Consider a spherical shell of radius $$R$$ at temperature $$T$$. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume $$u = \displaystyle \dfrac {U} {V} \propto T^4 $$ and pressure $$P = \displaystyle \dfrac {1} {3} \left( \frac {U} {V} \right) $$ . If the shell now undergoes an adiabatic expansion the relation between T and R is :

A
$$ T \propto e^{-R} $$

B
$$ T \propto e^{-3R} $$

C
$$\displaystyle T \propto \dfrac {1} {R} $$

D
$$\displaystyle T \propto \dfrac {1} {R^3} $$

Solution

$$\displaystyle \frac {U} {V} \propto T^4 $$ $$\Rightarrow U=CV{ T }^{ 4 }$$ where C is constant.
$$\displaystyle P=\frac { 1 }{ 3 } \left( \frac { U }{ V } \right) =\frac { 1 }{ 3 } \left( \frac { CV{ T }^{ 4 } }{ V } \right) =\left( \frac { C{ T }^{ 4 } }{ 3 } \right) $$

For adiabatic expansion, $$dQ=0 \Rightarrow dU=-dW$$
$$\Rightarrow d\left( CV{ T }^{ 4 } \right) =-PdV$$
$$\Rightarrow 4CV{ T }^{ 3 }dT+C{ T }^{ 4 }dV=\dfrac { -C{ T }^{ 4 } }{ 3 } dV$$
$$\Rightarrow 4VdT=-TdV-\dfrac { { T } }{ 3 } dV\\ 4VdT=-\dfrac { { 4 } }{ 3 } TdV\\ \dfrac { dT }{ T } =-\dfrac { dV }{ 3V } \\ \ln T=-\dfrac { 1 }{ 3 } \ln V+\ln{ C }^{ ' }\\ \ln T{ V }^{ 1/3 }=\ln{ C }^{ ' }\\ T{ V }^{ \frac { 1 }{ 3 } }={ C }^{ ' }\\ T{ \left( \dfrac { 4 }{ 3 } \pi { R }^{ 3 } \right) }^{ \dfrac { 1 }{ 3 } }={ C }^{ ' }$$
$$\Rightarrow TR=$$ constant
$$\Rightarrow T \propto \dfrac{1}{R}$$
Question 9
1 / -0

The equation of state for a gas is given by $$PV=nRT+\alpha V$$, where n is the number of moles and $$\alpha$$ is a positive constant. The initial temperature and pressure of one mole of the gas contained in a cylinder are $$T_o$$ and $$P_o$$ respectively. The work done by the gas when its temperature doubles isobarically will be

A
$$\dfrac {P_oT_oR}{P_o-\alpha}$$

B
$$\dfrac {P_oT_oR}{P_o+\alpha}$$

C
$$P_oT_oR \ln 2$$

D
$$P_oT_oR$$

Solution

Isobaric process means pressure will be constant. For one mole gas, $$n=1$$
If $$V_i$$ is the initial volume, $$P_0V_i=RT+\alpha V_i , V_i=\dfrac{RT_0}{P_0-\alpha}$$
If $$V_f$$ is the final volume, $$P_0V_f=R.2T_0+\alpha V_f , V_f=\dfrac{2RT_0}{P_0-\alpha}$$
The work done $$=\int^{V_f}_{V_i}P_0 dV=P_0[V_f-V_i]=\dfrac{P_0RT_0}{P_0-\alpha}$$
Question 10
1 / -0

Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm from $$20^oC$$ to $$90^oC$$. Work done by gas close to : (Gas constant R = 8.31 J / mol. K)

A
73 J

B
291 J

C
581 J

D
146 J

Solution

Work Done =$$ P\Delta V = nR \Delta T=1\times 8.31\times (90-20) = 581 J$$

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Thermodynamics Test - 9

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Thermodynamics Test - 9

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Question 1

$$200$$ $$\mathrm{R}$$

$$300$$ $$\mathrm{R}$$

$$400$$ $$\mathrm{R}$$

$$500$$ $$\mathrm{R}$$

Solution

Question 2

temperature

pressure

work

volume

Solution

Question 3

$$ \dfrac{4}{3}$$

$$2$$

$$\dfrac{5}{3}$$

$$\dfrac{3}{2}$$

Solution

Question 4

0.32

0.24

0.15

0.08

Solution

Question 5

$$\dfrac{2}{3}$$

$$\dfrac{3}{2}$$

$$\dfrac{3}{5}$$

$$\dfrac{2}{5}$$

Solution

Question 6

$$\displaystyle \dfrac{3 \gamma + 5}{6}$$

$$\displaystyle \dfrac{3 \gamma - 5}{6}$$

$$\displaystyle \dfrac{\gamma + 1}{2}$$

$$\displaystyle \dfrac{\gamma - 1}{2}$$

Solution

Question 7

second law of thermodynamics

conservation of momentum

conservation of mass

first law of thermodynamics.

Solution

Question 8

$$ T \propto e^{-R} $$

$$ T \propto e^{-3R} $$

$$\displaystyle T \propto \dfrac {1} {R} $$

$$\displaystyle T \propto \dfrac {1} {R^3} $$

Solution

Question 9

$$\dfrac {P_oT_oR}{P_o-\alpha}$$

$$\dfrac {P_oT_oR}{P_o+\alpha}$$

$$P_oT_oR \ln 2$$

$$P_oT_oR$$

Solution

Question 10

73 J

291 J

581 J

146 J

Solution

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