Thermodynamics Test

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Thermodynamics Test - 86

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Weekly Quiz Competition

Question 1
1 / -0

In a thermodynamic process a gas expands such that the heat transferred to the gas $$ (d Q)= $$ decrease in internal energy $$ (-d U) . $$ Find

The molar heat capacity

A
$$\dfrac{R}{1-\gamma}$$

B
$$\dfrac{2 R}{\gamma-1}$$

C
$$\dfrac{R}{2 \gamma-1}$$

D
$$\dfrac{R}{\gamma+2}$$
Question 2
1 / -0

A cyclic process $$ A B C A $$ is shown in the $$ V-T $$ diagram. Process on the $$ P $$ - $$ V $$ diagram is

A

B

C

D

Solution

From the given $$ V-T $$ diagram, we can see that in process $$ A B, V \propto T . $$ Therefore pressure is constant (as quantity of the gas remains same)
In process $$ B C, V= $$ constant and in process $$ C A, T= $$ constant
Therefore these processes are correctly represented on $$ P-V $$ diagram by graph (c)
Question 3
1 / -0

A cyclic process $$ A B C D $$ is shown in the following $$ P-V $$ diagram. Which of the following curves represents the same process?

A

B

C

D

Solution

$$ A B $$ is isobaric process, $$ B C $$ is isothermal process, $$ C D $$ is isochoric process and $$ D A $$ is isothermal process.
These processes are correctly represented by graph (a).
Question 4
1 / -0

Carbon monoxide is carried around a closed cyclic process $$ a b c, $$ in which $$ b c $$ is an isothermal process, as shown in Fig. 2.115. The gas absorbs 7000 J of heat as its temperature is increased from 300 K to 1000 K in going from $$ a $$ to $$ b $$. The quantity of heat ejected by the gas during the process $$ c a $$ is

A
$$4200 J$$

B
$$500 J$$

C
$$9000 J$$

D
$$9800 J$$

Solution

$$(\Delta Q)_{a b}=+7000=\mu C_{v}(1000-300) \quad\quad\quad(i)$$

For the process $$ c a $$

$$T_{a} =300 \mathrm{K}, \quad T_{c}=T_{b}=1000 \mathrm{K}$$

$$(\Delta Q)_{c a} =\mu C_{p}(300-1000)=-\mu C_{p} \times 700 $$

$$=-\mu\left(C_{v}+R\right) 700 \quad\quad\quad(ii)$$

For carbon monoxide:

$$\gamma=\dfrac{7}{5}$$

$$C_{v}=\dfrac{R}{\gamma-1}=\dfrac{R}{\dfrac{7}{5}-1}=\dfrac{5 R}{2} \quad\quad\quad(iii)$$

Hence, from Eq. (i)

$$7000=\mu \dfrac{5 R}{2} \times 700 \text { or } \mu R=\dfrac{20}{5}=4$$

$$(\Delta Q)_{c a}=-(7000+4 \times 700)=-9800 \mathrm{J}$$

Negative sign shows that heat is ejected.
Question 5
1 / -0

An ideal gas (1 mol, monatomic) is in the initial state $$ P $$ (see Fig. 2.111 ) on an isothermal $$ A $$ at temperature $$ T_{0} . $$ It is brought under a constant volume $$ \left(2 V_{0}\right) $$ process to $$ Q $$ which lies on an adiabatic $$ B $$ intersecting the isothermal $$ A $$ at $$ \left(P_{0}, V_{0}, T_{0}\right) . $$ The change in the internal energy of the gas during the process is (in terms of $$ T_{0} $$ ) $$ \left(2^{2 / 3}=\right.1.587)$$

A
$$2.3 T_{0} $$

B
$$-4.6 T_{0}$$

C
$$-2.3 T_{0}$$

D
$$4.6 T_{0} $$

Solution

Temperature at state $$ P=T_{0}, $$ since $$ P $$ lies on the isothermal of temperature $$ T_{0} $$. If $$ T $$ be the temperature at $$ Q $$, then for the adiabatic process $$ B, $$ we have, $$ T_{0} V_{0}^{\gamma-1}=T\left(2 V_{0}\right)^{\gamma-1} $$

$$T=\dfrac{T}{2^{\gamma-1}}=\dfrac{T_{0}}{2^{2 / 3}}$$

Change in the internal energy of the gas is

$$\Delta U= C_{V}\left(T-T_{0}\right)=\left(\dfrac{R}{\gamma-1}\right)\left(\dfrac{T_{0}}{2^{2 / 3}}-T_{0}\right)$$

$$=\dfrac{3 R T_{0}\left(1-2^{2 / 3}\right)}{2 \times 2^{2 / 3}}=-4.6 T_{0}$$
Question 6
1 / -0

A monatomic gas "undergoes a cycle consisting of two isothermals and two isobarics. The minimum and maximum temperatures of the gas during the cycle are $$ T_{1}=400 $$. K and $$ T_{2} =800 \mathrm{K}, $$ respectively, and the ratio of maximum to minimum volume is 4

The volume at $$ D $$ is

A
$$ 1.5 V_{0} $$

B
$$ 2 V_{0} $$

C
$$ 3 V_{0} $$

D
$$ 2.5 V_{0} $$

Solution
Question 7
1 / -0

Two samples $$ A $$ and $$ B $$ of a gas initially at the same pressure and temperature are compressed from volume $$ V $$ to $$\dfrac{ V }{2 }$$ ($$A $$ isothermally and $$ B $$ adiabatically). The final pressure of $$ A $$ is

A
greater than the final pressure of $$ B $$

B
equal to the final pressure of $$ B $$

C
less than the final pressure of $$ B $$

D
twice the final pressure of $$ B $$

Solution

For isothermal process

$$P_{1} V=P_{2}^{\prime} \dfrac{V}{2} \Rightarrow P_{2}^{\prime}=2 P_{1}$$

For adiabatic process

$$P_{1} V^{\gamma}=P_{2}\left(\dfrac{V}{2}\right)^{\gamma}$$

$$\Rightarrow P_{2}=2^{\gamma} P_{1}$$

since $$ \gamma>1, P_{2}>P_{2}^{\prime} $$
Question 8
1 / -0

Three processes compose a thermodynamics cycle shown in the P-V diagram. Process $$ 1 \rightarrow 2 $$ takes place at constant temperature , process $$ 2 \rightarrow 3 $$ takes place at constant volume and process $$ 3 \rightarrow 1$$ is adiabatic during the complete cycle.The total amount of work done in the cycle is $$10 J$$. During process $$ 2 \rightarrow 3 $$ the internal energy decreases by $$20 J$$ and during process $$ 3 \rightarrow 1 $$, $$20 J$$ of work done on the system. How much heat is added to the system during process $$ 1 \rightarrow 2 $$?

A
$$0$$

B
$$10 J$$

C
$$20 J$$

D
$$30 J$$

Solution
Question 9
1 / -0

A monatomic gas "undergoes a cycle consisting of two isothermals and two isobarics. The minimum and maximum temperatures of the gas during the cycle are $$ T_{1}=400 $$. K and $$ T_{2} =800 \mathrm{K}, $$ respectively, and the ratio of maximum to minimum volume is 4

The efficiency of the cycle is

A
$$ \dfrac{3 \ln 2}{5+4 \ln 2} \times 100 \% $$

B
$$ \dfrac{2 \ln 2}{5+4 \ln 2} \times 100 \% $$

C
$$ \dfrac{3 \ln 3}{5+4 \ln 2} \times 100 \% $$

D
$$ \dfrac{2 \ln 2}{3+4 \ln 2} \times 100 \% $$

Solution

Process $$ C B $$ is done at minimum temperature, while process $$ D A $$ is done at maximum temperature. Process $$ A \rightarrow B $$ (isobaric)

$$\dfrac{V_{A}}{V_{B}}=\dfrac{T_{A}}{T_{B}}=\dfrac{800}{400}=2$$

hence $$ V_{B}=2 V_{0} $$

$$ B \rightarrow C (\text { isothermal }) $$ $$ C \rightarrow D $$ (isobaric)
$$\dfrac{V_{D}}{V_{C}}=\dfrac{T_{D}}{T_{C}} \Rightarrow V_{D}=2 V_{0}$$

$$ D \rightarrow A $$ (isothermal)
$$W_{A B} =P\left(V_{B}-V_{A}\right)=n R \mathrm{D} T $$
$$=n R(-400)=-400 n R$$
$$W_{B C} =n R T \ln \dfrac{V_{C}}{V_{B}}=n R 400 \ln \dfrac{\mathrm{V}_{0}}{2 \mathrm{V}_{0}}$$
$$=-400 \mathrm{n} R \ln 2 $$
$$W_{C D} =P\left(V_{D}-V_{C}\right)=n R(400)=400 \mathrm{n} R$$
$$ W_{D A} =n R T \ln \dfrac{V_{A}}{V_{D}} $$
$$=n R \times 800 \times \ln \dfrac{2 V_{0}}{V_{0}}=800 n R \ln 2 $$
$$ \Delta W =400 n R \ln 2 $$ $$ \Delta Q $$ heat is extracted $$ Q_{C D}=n C_{P} \Delta T=n \dfrac{5}{2} R 400=1000 n R $$
$$=800 n R \ln 2=1000 n R+800 n R \ln 2$$
$$ \text { Efficiency } =\dfrac{\Delta W}{\Delta Q}$$
$$ \eta =\dfrac{2 \ln 2}{5+4 \ln 2} \times 100 \% $$
Question 10
1 / -0

Which of the following is slow process

A
Isothermal

B
Adiabatic

C
Isobaric

D
None of these

Solution

isothermal processes are necessarily slow as they require heat transfer to remain at the same temperature which is done by being in thermal equilibrium with some reservoir. A process will be isothermal only if it happens on timescales larger than the timescale required for effective heat transfer.