Thermodynamics Test

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Thermodynamics Test - 74

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Weekly Quiz Competition

Question 1
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In an adiabatic change, the pressure and temperature of a monoatomic gas are related with relation as $$P \propto T^C, $$ where $$C$$ is equal to :

A
$$\dfrac{5}{4}$$

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

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

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

Solution

Equation for adiabatic process is given by
$$PV ^\gamma =constant$$
Foradiabatic process, Poisson's equation is given by
$$PV^\gamma=constant$$ ----(1)

Ideal gas relation is
$$PV=RT$$
$$\Rightarrow V=\dfrac{RT}{P}$$ ----(2)

From equations (1) and (2), we get
$$P\left ( \dfrac{RT}{P} \right )^\gamma=constant$$
$$\Rightarrow \dfrac{T^\gamma}{P^{\gamma-1}}=constant$$ ----(3)
where $$\gamma$$ is ration of specific heats of the gas.

Given, $$P \ \alpha \ T^C$$ ----(4)
On comparing with equation (3), we have
$$C=\dfrac{\gamma}{\gamma-1}$$
For a monoatomic gas $$\gamma =\frac{5}{3}$$
$$\therefore $$ We have $$\displaystyle C=\dfrac{\dfrac{5}{3}}{\dfrac{5}{3}-1}=\dfrac{5}{2}$$
Question 2
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When 100 J of heat is given to an ideal gas it expands from 200 $$cm^3$$ to 400 $$ cm^3$$ at a constant pressure of $$3 \times 10^5$$ Pa. Then calculate the change in internal energy of the gas :
$$R=\left [ \dfrac{25}{3}J/mol-k \right ]$$

A
10 J

B
40 J

C
30 J

D
None of these

Solution

$$\Delta U = \Delta Q - W$$ where $$\Delta U$$ is the change in internal energy, $$\Delta Q$$ is the heat supplied to the system and $$W$$ is the work done by the syatem.
Given: $$\Delta Q=100J$$
$$W= \int_{200}^{400} P dv = \int_{200}^{400} 3 \times 10^5 dv= 3 \times 10^5 \times (400-200) \times 10^{-6}=60\:J$$
$$\Delta U = \Delta Q - W= (100 -60)=40\:J$$
Question 3
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Which of the following represents correctly the changes in thermodynamics properties during the formation of 1 mol of an ideal binary solution?

A

B

C

D

Solution
Question 4
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Net heat released by the system if initial and final temperatures of a gas is same and work done is $$35 kJ$$ is

A
positive

B
negative

C
zero

D
cannot be decided with given information

Solution

From 1st Law,
$$dQ=dU +dW$$
Since initial and final temps are same, $$dU=0$$
Given,
$$dW= - 35\:J$$
$$ \therefore $$ $$dQ= 0 - 35\:J = -35 \ J$$
$$\therefore$$ $$35$$ J heat is released by the system.
Question 5
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A mass of diatomic gas ($$\gamma = 1.4$$) at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rise from 27$$^o$$C to 927$$^o$$C. The pressure of the gas is final state is

A
28 atm

B
68.7atm

C
256 atm

D
8 atm

Solution

For an adiabatic process
$$\frac{T^{\gamma}}{P^{\gamma-1}}=constant$$
Given,
$$T_1=27^o\:C=273+27=300^o\:K$$
$$T_2=927^o\:C=273+927=1200^o\:K$$
$$P_1=2\: atm$$
$$\gamma=1.4$$

$$\dfrac{T_1^{\gamma}}{P_1^{\gamma-1}}=\dfrac{T_2^{\gamma}}{P_2^{\gamma-1}}$$

Substituting the values,

$$\dfrac{300^{1.4}}{2^{1.4-1}}=\dfrac{1200^{1.4}}{P_2^{1.4-1}}$$

$$ \therefore (\dfrac {P_2}{2})^{0.4}=(\dfrac{1200}{300})^{1.4}$$

$$ \ln ((\dfrac {P_2}{2})^{0.4}= \ln (\dfrac{1200}{300})^{1.4}$$

$$\therefore 0.4 \ln (\dfrac{P_2}{2}) = 1.4 \ln(\dfrac{4}{1})= 1.4^2$$

$$\therefore \ln (\dfrac{P_2}{2})= 4.9$$

$$\therefore P_2=256\:atm$$
Question 6
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Directions For Questions

Two moles of an ideal gas at temperature $$T_o=300\:K$$ was cooled isochorically so that the pressure was reduced to half. Then, in an isobaric process, the gas expanded till its temperature got back to the initial value.

...view full instructions

Find work done in the isobaric process.

A
$$W=1245\:J$$

B
$$W=1255\:J$$

C
$$W=1254\:J$$

D
$$W=1240\:J$$

Solution

Isochoric process:
$$\Delta W=0$$ as $$\Delta V=0$$
$$\Delta U=C_v \Delta T=\dfrac{R}{\gamma -1} \times \Delta T$$
$$ \Rightarrow \Delta U = \dfrac{\Delta(RT)}{\gamma-1}$$

$$ \Rightarrow \dfrac{V}{\gamma -1} \Delta P=-\dfrac{1}{2} \dfrac{pV}{\gamma-1}$$ since $$\Delta P= \dfrac{p}{2}-p=-\dfrac{p}{2}$$
$$\Delta Q= \Delta U + \Delta W$$
$$

\therefore \Delta Q= -\dfrac{1}{2} \dfrac{pV}{\gamma-1} +0 = -\dfrac{1}{2}

\dfrac{pV}{\gamma-1}= -\dfrac{1}{2} \dfrac{RT_0}{\gamma-1}$$

In the isochoric process, temp $$T_0$$ is reduced to half i.e to $$\dfrac{T_0}{2}$$ since $$P \ \alpha \ T$$
Isobaric process:
$$ \Delta W = \int_{V_i}^{V_f}P dV=\dfrac {P}{2}(V_f-V_i)$$
Since the original temp is restored to $$T_0$$, $$\Delta T = \dfrac{T_0}{2}$$

$$ \therefore \Delta W= \dfrac{1}{2} R T_0= \dfrac{1}{2} \times 8.3 \times 300=1245\:J$$
Question 7
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Piston cylinder device initially contains $$0.5 m^3$$ of nitrogen gas at $$400 kPa$$ and $$27^o$$C. An electric heater within the device is at turned on and is allowed to pass a current of $$2 A$$ for $$5 \ min$$ from a $$120 V$$ source. Nitrogen expands at constant pressure and a heat loss of $$2800 J$$ occurs during the process.
$$R = \dfrac{25}{3} kJ/kmol-K$$
The final temperature of nitrogen is:

A
$$2.6^oC$$

B
$$56.6^oC$$

C
$$29.6^oC$$

D
$$5.67^oC$$

Solution

Heat supplied to the gas by the heater $$H= 120 \times 2 \times (5 \times 60) = 72 \:kJ$$
Heat lost: $$H_{lost}= 2800\:J=2.8 \:kJ$$
mass of nitrogen $$m=\dfrac{P_1V_1}{RT_1}=\dfrac{ 400\: kPa \times 0.5 m^3}{0.297 \ kPa m^3 kg^{-1} K^{-1} }\times ( 27 +273)\:K= 2.245 \:kg$$ where $$R=0.297 \ kPa\: m^3 kg^{-1} K^{-1} $$

Work done by the gas at constant Pressure $$W= H- H_{lost} = 72-2.8 = 69.2\: J$$
$$ \therefore mC_p(T_2- 300) =69.2$$
$$ \therefore 2.245 \times 1.039( T_2- 300)=69.2$$
$$\therefore T_2= 56.7^o\:C$$
Question 8
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A gas can expand through two processes : (i) isobaric, (ii) P/V = constant. Assuming that the initial volume is same in both processes and the final volume which is two times the initial volume is also same in both processes, which of the following is true ?

A
Work done by gas in process (i) is greater than the work done by the gas in process (ii)

B
Work done by gas in process (i) is smaller than the work done by the gas in process (ii)

C
Final pressure is greater in process (i)

D
Final temperature is greater in process (i)

Solution

For Isobaric Process Work Done is given by: $$W={ P }({ V }_{ 2 }-{ V }_{ 1 })$$
So Work Done in isobaric:
$${ W }_{ iso }\quad =\quad { P }_{ i }({ V }_{ f }-{ V }_{ i })\quad =\quad { P }_{ i }({ 2V }_{ i }-{ V }_{ i })\quad =\quad { P }_{ i }{ V }_{ i }$$

Work Done in P/V= constant is given by:
$${ W }_{ P/V=Con }\quad =\quad \int _{ { V }_{ i } }^{ { V }_{ f } }{ PdV } =\quad \int _{ { V }_{ i } }^{ { V }_{ f } }{ cVdV } =\quad c{ [(2{ V }_{ i }) }^{ 2 }-({ V }_{ i })^{ 2 }]/2=3c{ (V }_{ i }^{ 2 })/2=\dfrac { 3 }{ 2 } \dfrac { { P }_{ i } }{ { V }_{ i } } { { V }_{ i } }^{ 2 }\quad =\quad \dfrac { 3 }{ 2 } { P }_{ i }{ V }_{ i }$$

Hence, Work done by gas in process (i) is smaller than the work done by the gas in process (ii)
Question 9
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Directions For Questions

One mole of an ideal gas is taken from an initial state $$P_o, V_o$$ and temperature $$T_o$$ through the following activities. ($$\gamma = \dfrac{C_p}{C_v}, C_p = \dfrac{7R}{2}, C_v = \dfrac{5R}{2}$$)
(i) Heating at constant volume to a temperature three times.
(ii) Adiabatic expansion to a volume $$2V_o$$
(iii) Cooling at constant volume to a temperature on third
(iv) Adiabatic compression so that it is returned to its initial state

...view full instructions

The temperature at D is

A
$$T_o 2^{1.4}$$

B
$$T_o 2^{0.4}$$

C
$$T_o 2^{-1.4}$$

D
$$T_o 2^{-0.4}$$

Solution

$$\gamma = \dfrac{\dfrac{7R}{2}}{\dfrac{5R}{2}} = 1.4$$

Heating at a constant volume so final temprature $$T=3T_o$$
So, we can calculate final pressure $$P = 3P_o$$

Adiabatic expansion
$$P_iV_i^\gamma =P_fV_f^\gamma $$
$$P_f = \dfrac{3P_0}{2^{1.4}}$$

$$T_f= \dfrac{3T_o}{2^{0.4}}$$

Cooling at constant volume with temp one third of previous state, hence
$$T = \dfrac{T_o}{2^{0.4}}$$
Question 10
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A sound wave passing through air at $$NTP$$ produces a pressure of $$0.001\ dyne/cm^2$$ during a compression. The corresponding change in temperature (given $$\gamma = 1.5$$ and assume gas to be ideal) is

A
$$8.97 \times 10 ^{-4}K$$

B
$$8.97 \times 10 ^{-6}K$$

C
$$8.97 \times 10 ^{-8}K$$

D
none of these

Solution

T$$^\gamma$$ P$$^{1-\gamma}$$ = constant.
Differentiating above equation
$$\displaystyle \gamma T^{\gamma-1} dT P^{1-\gamma} + T^\gamma (1-\gamma) P^{-\gamma} dP = 0$$
or, $$ dT = \displaystyle \cfrac{(\gamma-1)T}{\gamma P}dP$$
or, $$dT =\displaystyle \left( \cfrac { 1.5-1 }{1.5 } \right) \left( \cfrac { 273 }{76 \times 13.6 \times 981 } \times 0.001\right) $$
= $$8.97 \times 10^{-8}K$$