Thermodynamics Test

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Thermodynamics Test - 35

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Question 1
1 / -0

A thermodynamics cycle takes in heat energy at a high temperature and rejects energy at a lower temperature. If the amount of energy rejected at the low temperature is 3 times the amount of work done by the cycle, the efficiency of the cycle is:

A
$$0.25$$

B
$$0.33$$

C
$$0.67$$

D
$$0.9$$

Solution

Let heat taken be $$ Q_1$$, heat rejected be $$Q_2$$, and work done be W
Then $$Q_2=3W$$
Also, $$Q_1=Q_2 + W$$
$$\Rightarrow Q_1=4W$$
Efficiency of cycle=work done/heat taken=$$\dfrac{W}{4W}=\dfrac{1}{4}=0.25$$
Question 2
1 / -0

Monoatomic , diatomic and triatomic gases whose initial volume and pressure are same , are compressed till their volume becomes half the initial volume.

A
If the compression is adiabatic then the monoatomic gas will have maximum final pressure

B
If the compression is adiabatic then triatomic gas will have maximum final pressure

C
If the compression is adiabatic then their final pressure will be same.

D
If the compression is isothermic then final pressure will be different

Solution

For an adiabatic process,
$$ {P} {V}^{\gamma} = constant $$
or
$$\dfrac{P_1}{P_2}=(\dfrac{V_2}{V_1})^{\gamma}$$
or
$$P_2=2^{\gamma}P_1$$
Now, for a monoatomic gas, the value of $$ \gamma $$ is the highest
Thus, for the same change in volume, the monoatomic gas will have the maximum pressure.
Question 3
1 / -0

A reversible adiabatic path on a P-V diagram for an ideal gas passes through sate A where $$ P=0.7 \times 10^5 N/{m}^{-2} $$ and $$v=0.0049\ m^3$$. The ratio of specific heat of the gas is 1.4 . The slope of path at A is:

A
$$2.0 \times 10^7 N{m}^{-5} $$

B
$$1.0 \times 10^7 N{m}^{-5} $$

C
$$-2.0 \times 10^7 N{m}^{-5} $$

D
$$-1.0 \times 10^7 N{m}^{-5} $$

Solution

For reversible adiabat,
$$Pv^\gamma=constant \Rightarrow vdP + P\gamma dv=0 \Rightarrow \dfrac{dP}{dv}=-\dfrac{\gamma P}{v}$$
For $$P=0.7 \times 10^5Nm^{-2}, v=0.0049m^3, \gamma =1.4$$
required slope= $$-\dfrac{1.4\times 0.7\times 10^5\ Nm^{-2}}{0.0049\ m^3}=-2\times 10^7Nm^{-5}$$
Question 4
1 / -0

$$1\ kg $$ of gas does $$20\ kJ$$ of work and receives $$16\ kJ$$ of heat when it is expanded between two states. A second kind of expansion can be found between the initial and final state which requires a heat input of $$9\ kJ$$. The work done by the gas in the second expansion is:

A
$$32\ kJ$$

B
$$5\ kJ$$

C
$$-4 kJ$$

D
$$13\ kJ$$

Solution

The change in the internal energy will remain the same, since the initial and final states are the same for both the processes.
Thus, U = Q - W
16 - 20 = 9 - W
$$W = 13\ kJ$$
Question 5
1 / -0

On an isothermal process, there are two POINTs $$A$$ and $$B$$ at which pressures and volumes are $$(2P_o,\:V_o)$$ and $$(P_o,\:2V_o)$$ respectively. If $$A$$ and $$B$$ are connected by a straight line, the temperature of the isothermal process is how much lower the maximum temperature

A
$$\displaystyle\frac{P_oV_o}{nR}$$

B
$$\displaystyle\frac{P_oV_o}{2nR}$$

C
$$\displaystyle\frac{3P_oV_o}{2nR}$$

D
$$\displaystyle\frac{P_oV_o}{4nR}$$

Solution

Using symmetry, the temperature will be maximum at a point where the pressure and the volume attain the same value.
Therefore, on a straight line, joining the 2 points A and B as given in the question, the temperature will be maximum at their midpoint.
Hence, the pressure will be $$ \dfrac{3 {P}_{o}}{2} $$ and the volume will be $$ \dfrac{3 {V}_{o}}{2} $$
Thus, for n moles of a gas, the maximum temperature is,
$$ T = \dfrac{9 {P}_{o} {V}_{o}}{4nR} $$
However, the isothermal temperature is,
$$ T = \dfrac{2 {P}_{o} {V}_{o}}{nR} $$
Thus, the difference in 2 temperatures will be,
$$ \dfrac{{P}_{o} {V}_{o}}{4nR} $$
Question 6
1 / -0

A vessel contains an ideal monoatomic gas which expands at constant pressure, when heat $$Q$$ is given to it. What is the work done?

A
$$Q$$

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

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

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

Solution

For monoatomic gas,$$ C_p=\dfrac{5R}{2}, C_v=\dfrac{3R}{2}$$
Hence for isobaric process, heat given=$$Q=\dfrac{5}{2}nR\Delta T=\dfrac{3}{2}nR\Delta T + W$$
$$\Rightarrow W=nR\Delta T=\dfrac{2}{5}Q$$
Question 7
1 / -0

In an adiabatic process, $$\displaystyle R=\frac{2}{3}C_v$$. The pressure of the gas will be proportional to:

A
$$T^{5/3}$$

B
$$T^{5/2}$$

C
$$T^{5/4}$$

D
$$T^{5/6}$$

Solution

It is given that, $$ {C}_{v} = 1.5 R $$
Using, $$ {C}_{p} = {C}_{v} + R $$,
$$ {C}_{p} = 2.5 R $$
Thus, $$ \gamma = \dfrac{2.5}{1.5} = 5/3$$

Now,
Using the condition of adiabaticity,
$$ {P}^{1 - \gamma} {T}^{\gamma} = constant $$
Substituting the value of $$ \gamma $$ ,
Pressure is directly proportional to $$ {T}^{5/2} $$
Question 8
1 / -0

An ideal gas confined into an insulated chamber is allowed to enter into an evacuated insulated chamber. If Q,W and $$ \Delta E _{int} $$ have the usuak meanings, then

A
Q=0, W $$ \neq 0$$

B
W=0, Q $$ \neq 0$$

C
$$ \Delta E _{int} =0, Q \neq 0$$

D
$$Q=W, \Delta E _{int} =0$$

Solution
Question 9
1 / -0

Certain perfect gas is found to obey $$ P{V} ^{3/2} = constant $$ during an adiabatic process. If such a gas at initial temperature T is adiabatically compressed to half the initial volume, its final temperature will be

A
$$\sqrt{2}T$$

B
$$2T$$

C
$$2\sqrt{2} T$$

D
$$4T$$

Solution

Using the ideal gas law, P is directly proportional to $$ \dfrac{T}{V} $$
Substituting this value,
$$ T {V}^{0.5} = constant $$
Thus, when the volume is halved, we raise the temperature by $$ \sqrt 2 $$ times.
Question 10
1 / -0

One mol of a gas at NTP is suddenly expanded to three times its initial volume. If $$C_v =2 R$$, the ratio of initial to final pressure of gas will nearly be

A
$$5$$

B
$$4$$

C
$$3$$

D
$$2$$

Solution

$$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$$
$$P_2 = P_1 \displaystyle \left ( \dfrac{V_1}{V_2} \right )^{\gamma} = 1 \left ( \dfrac{1}{3} \right )^{3/2}$$
$$\displaystyle \dfrac{P_1}{P_2} = 3 \sqrt{3} =5.1$$

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Thermodynamics Test - 35

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Thermodynamics Test - 35

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Question 1

$$0.25$$

$$0.33$$

$$0.67$$

$$0.9$$

Solution

Question 2

If the compression is adiabatic then the monoatomic gas will have maximum final pressure

If the compression is adiabatic then triatomic gas will have maximum final pressure

If the compression is adiabatic then their final pressure will be same.

If the compression is isothermic then final pressure will be different

Solution

Question 3

$$2.0 \times 10^7 N{m}^{-5} $$

$$1.0 \times 10^7 N{m}^{-5} $$

$$-2.0 \times 10^7 N{m}^{-5} $$

$$-1.0 \times 10^7 N{m}^{-5} $$

Solution

Question 4

$$32\ kJ$$

$$5\ kJ$$

$$-4 kJ$$

$$13\ kJ$$

Solution

Question 5

$$\displaystyle\frac{P_oV_o}{nR}$$

$$\displaystyle\frac{P_oV_o}{2nR}$$

$$\displaystyle\frac{3P_oV_o}{2nR}$$

$$\displaystyle\frac{P_oV_o}{4nR}$$

Solution

Question 6

$$Q$$

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

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

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

Solution

Question 7

$$T^{5/3}$$

$$T^{5/2}$$

$$T^{5/4}$$

$$T^{5/6}$$

Solution

Question 8

Q=0, W $$ \neq 0$$

W=0, Q $$ \neq 0$$

$$ \Delta E _{int} =0, Q \neq 0$$

$$Q=W, \Delta E _{int} =0$$

Solution

Question 9

$$\sqrt{2}T$$

$$2T$$

$$2\sqrt{2} T$$

$$4T$$

Solution

Question 10

$$5$$

$$4$$

$$3$$

$$2$$

Solution

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