Thermodynamics Test

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Thermodynamics Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

A diatomic gas follows equation $$ PV^m = constant $$ during a process.what should be value of m such that its molar heat capacity during process = R?

A
2/3

B
1

C
1.5

D
5/3

Solution
Question 2
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 $$ B $$ is

A
$$ 1.5 V_{0} $$

B
$$ 2 V_{0} $$

C
$$ 3 V_{0} $$

D
$$ 2.5 V_{0} $$

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 $$ V_{B}=2 V_{0} $$
Question 3
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A sample of ideal gas is expanded to twice its original volume of $$ 1 \mathrm{m}^{3} $$ in a quasi-static process for which $$ P= \alpha V^{2} $$ with $$ \alpha=3 \times 10^{5} \mathrm{Pa} / \mathrm{m}^{6} $$ as shown in Fig. $$ 2.113 . $$ Work done by the expanding gas is

A
$$8 \times 10^{5} \mathrm{J}$$

B
$$7 \times 10^{5} \mathrm{J}$$

C
$$6 \times 10^{5} \mathrm{J}$$

D
$$3 \times 10^{5} \mathrm{J}$$

Solution

Work done, $$ W=\int_{V_{1}}^{v_{2}} \alpha V^{2} d V $$

where $$ \alpha=3 \times 10^{5} \mathrm{Pa} / \mathrm{m}^{6} $$

$$ \Rightarrow \quad W=\left[\dfrac{\alpha V^{3}}{3}\right]_{V_{1}=1 \mathrm{m}^{3}}^{V_{2}=2 \mathrm{m}^{3}}=\left[\dfrac{3 \times 10^{5} V^{3}}{3}\right]_{V_{1}}^{V_{2}} $$

$$ \Rightarrow \quad W=10^{5}\left(V_{2}^{3}-V_{1}^{3}\right) $$

$$ V_{1}=1 \mathrm{m}^{3} \quad V_{2}=2 \mathrm{m}^{3} $$

$$ W=10^{5}(8-1)=7 \times 10^{5} \mathrm{J} $$
Question 4
1 / -0

A cyclic process for 1 mole of an ideal gas is shown in Fig. 2.98 in the $$ V-T, $$ diagram. The work done in $$ A B, B C $$ and $$ C A $$, respectively, is

A
$$ 0, R T_{2} \ln \left(\dfrac{V_{1}}{V_{2}}\right), R\left(T_{1}-T_{2}\right) $$

B
$$ R\left(T_{1}-T_{2}\right), 0, R T_{1} \ln \left(\dfrac{V_{1}}{V_{2}}\right) $$

C
$$ 0, R T_{2} \ln \left(\dfrac{V_{2}}{V_{1}}\right), R\left(T_{1}-T_{2}\right) $$

D
$$ 0, R T_{2} \ln \left(\dfrac{V_{2}}{V_{1}}\right), R\left(T_{2}-T_{1}\right) $$

Solution

Process $$ A B $$ is isochoric; therefore

$$W_{A B}=P \Delta V=0$$ Process $$ B C $$ is isothermal; therefore

$$W_{B C}=R T_{2} \ln \left(\dfrac{V_{2}}{V_{1}}\right)$$

Process $$ C A $$ is isobaric; therefore

$$W_{C A}=P \Delta V=R \Delta T=R\left(T_{2}-T_{1}\right)$$
Question 5
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 heat is extracted from the system in process

A
$$ A \rightarrow B, B \rightarrow C $$

B
$$ C \rightarrow D, D \rightarrow A $$

C
$$ A \rightarrow B, C \rightarrow D $$

D
$$ B \rightarrow C, C \rightarrow D $$

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)

So, option $$A$$ is correct.
Question 6
1 / -0

An ideal gas is taken through $$ A \rightarrow B \rightarrow C \rightarrow A $$, as shown in Fig. $$ 2.117 . $$ If the net heat supplied to the gas in the cycle is $$ 55 \mathrm{J} $$, the work done by the gas in the process $$ C \rightarrow A $$ is

A
$$-5 J$$

B
$$-10 J$$

C
$$-15 J$$

D
$$-20 J$$

Solution

$$d U=0 $$ Therefore by the first law of thermodynamics

$$ d Q_{\text {Cyclic }}=d W_{\text {Cyclic }} $$ since $$ B \rightarrow C $$ is an isochoric process

$$\Rightarrow d W_{a \rightarrow c}=0 $$

$$\Rightarrow 5=d W_{A \rightarrow B}+d W_{a \rightarrow C}+d W_{C \rightarrow A}$$

$$\Rightarrow 5=10(2-1)+0+d W_{c \rightarrow A}$$

$$\Rightarrow d W_{C \rightarrow A}=-5 \mathrm{J}$$
Question 7
1 / -0

One mole of an ideal monatomic gas undergoes thermodynamic cycle $$ 1 \rightarrow 2 \rightarrow 3 \rightarrow 1 $$ as shown in Fig. $$ 2.161 . $$ Initial temperature of gas is $$ T_{0}=300 \mathrm{K} $$
Process $$ 1 \rightarrow 2: P=a V $$
Process $$ 2 \rightarrow 3: P V= $$ Constant
Process $$ 3 \rightarrow 1: P= $$ Constant
(Take $$ \ln |3|=1.09) $$

Determine the heat capacity of each process

A
$$ 20.75 \mathrm{J} / \mathrm{mol}-\mathrm{K} $$

B
$$ 10.23 \mathrm{J} / \mathrm{mol}-\mathrm{K} $$

C
$$ 22.37 \mathrm{J} / \mathrm{mol}-\mathrm{K} $$

D
$$ 15.96 \mathrm{J} / \mathrm{mol}-\mathrm{K} $$

Solution

For process $$ 1 \rightarrow 2 $$
$$W_{12} =\int_{1}^{2} \alpha V d V=\alpha \int_{V_{0}}^{3 v_{0}} V d V=\dfrac{\alpha}{2}\left(9 V_{0}^{2}-V_{0}^{2}\right)$$

$$=4 \alpha V_{0}^{2}$$
Using gas law, $$ \dfrac{P_{1} V_{1}}{T_{1}}=\dfrac{P_{2} V_{2}}{T_{2}} $$

$$T_{2}=\dfrac{P_{2} V_{2}}{P_{1} V_{1}} T_{1}=\dfrac{V_{2}^{2}}{V_{1}^{2}} T_{1}=\left(\dfrac{3 V_{0}}{V_{0}}\right)^{2} T_{0}=9 T_{0}$$

For process $$ 2 \rightarrow 3 $$

$$W_{23} =R T_{2} \log \left|\dfrac{P_{2}}{P_{3}}\right|=R\left(9 T_{0}\right) \log \left|\dfrac{3 P_{0}}{P_{0}}\right| $$

$$=9 R T_{0} \log |3|=9.81 R T_{0}$$

For isothermal process: $$ P_{2} V_{2}=P_{3} V_{3} $$

Therefore,$$V_{3}=\dfrac{P_{2} V_{2}}{P_{3}}=\frac{3 P_{0}}{P_{0}}\left(3 V_{0}\right)=9 V_{0}$$

Also, $$ W_{31}=P_{0}\left(V_{1}-V_{3}\right)=P_{0}\left(V_{0}-9 V_{0}\right)=-8 P_{0} V_{0} $$ $$ =-8 R T_{0} $$

Applying gas law in process $$ 1 \rightarrow 2 $$

$$P_{0} V_{0}=R \mathrm{T}_{0} \quad \text { or } \quad \alpha V_{0}^{2}=R T_{0}$$

The net work is $$W_{\text {net }} =W_{12}+W_{23}+W_{31} $$

$$=4 R T_{0}+9.81 R T_{0}-8 R T_{0} $$

$$=5.81 R T_{0}$$

For process $$ 1 \rightarrow 2 $$
$$\Delta U_{12} =C_{V}\left(T_{2}-T_{1}\right)$$

$$=4 R T_{0}+12 R T_{0}=16 R T_{0}$$

since,$$\Delta U_{12} =C_{12}\left(T_{2}-T_{1}\right)=8 C_{12} T_{0}$$

$$C_{12} =2 R=16.6 \mathrm{J} / \mathrm{mol}-\mathrm{K}$$

For the process $$ 2 \rightarrow 3: C_{23}=\infty $$ For the process $$ 3 \rightarrow 1: \mathrm{C}_{31}=C_{\mathrm{P}}+R =5 R / 2=20.75 \mathrm{J} / \mathrm{mol}-\mathrm{K}$$
Question 8
1 / -0

Variation of internal energy with density of 1 mole of monatomic gas is depicted in Fig. 2.107 . Corresponding variation of pressure with volume can be depicted as (assume the curve is rectangular hyperbola)

A

B

C

D

Solution

Since $$ U \rho= $$ constant
$$\dfrac{P}{\rho}=\dfrac{R T}{M}$$
$$ P= $$ constant since $$ \rho $$ is increasing, therefore $$ V $$ is decreasing.
Question 9
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A fixed mass of gas is taken through a process $$ A \rightarrow B \rightarrow C \rightarrow A $$
$$ A \rightarrow B $$ is isobaric, $$ B \rightarrow C $$ is adiabatic and $$ C \rightarrow A $$ is isothermal.
Find pressure at $$ C $$

A
$$\dfrac{10^{5}}{64} \mathrm{N} / \mathrm{m}^{2}$$

B
$$\dfrac{10^{5}}{32} \mathrm{N} / \mathrm{m}^{2}$$

C
$$\dfrac{10^{5}}{12} \mathrm{N} / \mathrm{m}^{2}$$

D
$$\dfrac{10^{5}}{6} \mathrm{N} / \mathrm{m}^{2}$$

Solution

(a) For adiabatic process BC,
$$P_B = P_C$$ (i)

For isothermal process CA,
$$P_AV_A = P_CV_C$$ (ii)

From Eq (i) & (ii)

$$V_c = \left[\dfrac{V_B^{Y}}{V_A}\right]^{\dfrac{1}{\gamma -1}} = 64 m^3$$

$$P_c = \dfrac{P_A V_A}{V_C} = \dfrac{10^5}{64} N/m^2$$

(b) Work done, $$W = W_{AB} + W{BC} + W{CA}$$

$$= P(V_A - V_A) + \dfrac{1}{\gamma - 1}[PV_B - P_CV_C + PV_A In (V_A/V_C)]$$

Putting the values,

$$W = 4.9 \times 10^5 J$$
Question 10
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A fixed mass of gas is taken through a process $$ A \rightarrow B \rightarrow C \rightarrow A$$
$$ A \rightarrow B $$ is isobaric, $$ B \rightarrow C $$ is adiabatic and $$ C \rightarrow A $$ is isothermal.
Find volume at C

A
$$32 \mathrm{m}^{3}$$

B
$$100 \mathrm{m}^{3}$$

C
$$64 \mathrm{m}^{3}$$

D
$$25 \mathrm{m}^{3}$$

Solution

For adiabatic process BC,
$$P_B = P_C$$ (i)

For isothermal process CA,
$$P_AV_A = P_CV_C$$ (ii)

From Eq (i) & (ii)

$$V_c = \left[\dfrac{V_B^{Y}}{V_A}\right]^{\dfrac{1}{\gamma -1}} = 64 m^3$$