Thermodynamics Test

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Thermodynamics Test - 34

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Question 1
1 / -0

Two samples $$A$$ and $$B$$ of a gas initially at the same pressure and temperature are compressed from volume $$V$$ to $$\displaystyle\frac{V}{2}$$ ($$A$$ 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, $$PV =$$ constant
$$P_1V_1 = P_2V_2$$
$$\implies P_2 = 2P_1$$

For adiabatic process, $$ P {V}^{\gamma} $$ = constant
$$P_1V_1^{\gamma} = P_2V_2^{\gamma}$$
$$\implies P_2 = P_1 (\dfrac{V_1}{V_2})^{\gamma}$$
$$ \implies P_2 = P_1 2^{\gamma}$$

Since, $$ \gamma $$ is always greater than 1, For the same change in volume, there will be a greater change in pressure for adiabatic process than that for isothemral process.
Question 2
1 / -0

Four curves $$A$$, $$B$$, $$C$$ and $$D$$ are drawn in figure for a given amount of gas. The curves which represent adiabatic and isothermal changes are.

A
$$C$$ and $$D$$, respectively

B
$$D$$ and $$C$$, respectively

C
$$A$$ and $$B$$, respectively

D
$$B$$ and $$A$$, respectively

Solution

The PV graph is always a rectangular hyperbola with its asymptotes as the positive x and y axes.
However, $$the \ slope\ of\ the\ adiabatic\ curve\ is\ \gamma \ times\ the\ slope\ for\ an\ isothermal\ curve.$$

Thus, A curve is for adiabatic and B for isothermal.
Question 3
1 / -0

Directions For Questions

The temperature of $$3\:kg$$ of nitrogen is raised from $$10^\circ C$$ to $$100^\circ C$$ at constant pressure.
For nitrogen $$C_p=1400\:J/kgK$$ and $$C_v=740\:J/kgK$$

...view full instructions

Compute the change in internal energy (in joules)

A
$$199800$$

B
$$0$$

C
$$200000$$

D
$$200200$$

Solution

The change in internal energy is given by,
U = $$ m {C}_{v} \Delta T $$(m is used because $$C_p$$ and $$C_v$$ are given per kg)
U = 3 $$ \times 740 \times (100 - 10) $$
$$U = 199800 J$$
Question 4
1 / -0

Directions For Questions

The temperature of $$3\ moles$$ of nitrogen is raised from $$10^\circ C$$ to $$100^\circ C$$ at constant volume.
For nitrogen $$C_p=1400\:J/kgK$$ and $$C_v=740\:J/kgK$$

...view full instructions

Compute the work done (in joules)

A
$$199800$$

B
$$0$$

C
$$200000$$

D
$$200200$$

Solution

For the processes taking place at constant volume, the work done is always zero. Hence B is correct answer.
Question 5
1 / -0

A container of volume $$1\:m^3$$ is divided into two equal compartments by a partition. One of these compartments contains an ideal gas at $$300\:K$$. The other compartment is vacuum. The whole system is thermally isolated from its surroundings. The partition is removed and the gas expands to occupy the whole volume of the container. Its temperature now would be :

A
$$300\:K$$

B
$$250\:K$$

C
$$200\:K$$

D
$$100\:K$$

Solution
Question 6
1 / -0

An ideal gas $$\displaystyle\left(\frac{C_p}{C_v}=\gamma\right)$$ is taken through a process in which the pressure and the volume vary as $$P=aV^b$$. The value of b for specific heat capacity as zero is

A
$$2 \gamma$$

B
$$0$$

C
$$-\gamma$$

D
$$-1$$

Solution

Given that $$P=aV^b$$
or $$PV^{-b}=a$$
Comparing with $$PV^r=constant$$, we have
$$r=-b$$
We know that molar specific heat $$\displaystyle C=C_v-\dfrac{R}{r-1}$$

$$\displaystyle\therefore 0=\dfrac{R}{\gamma-1}-\dfrac{R}{-b-1}$$

or $$b=-\gamma$$
Question 7
1 / -0

A thermally insulated container is divided into two parts by a screen. In one part the pressure and temperature are $$P$$ and $$T$$ for an ideal gas filled. In the second part it is vacuum. If now a small hole is created in the screen, then the temperature of the gas will

A
decrease

B
increase

C
remain same

D
none of these

Solution

According to the ideal gas law, PV = nRT,
Since the will be a change in volume of gas occurs slowly i.e. process occurs isothermally.
(To compensate for the change in volume, there will be a corresponding change in the pressure of the gas, therefore the temperature will tend to remain constant.)
Question 8
1 / -0

In an adiabatic process pressure is increased by $$\displaystyle\frac{2}{3}\%$$ if $$\displaystyle\frac{C_p}{C_v}=\frac{3}{2}$$. Then the volume decrease by about

A
$$\displaystyle\frac{4}{9}\%$$

B
$$\displaystyle\frac{2}{3}\%$$

C
$$4\%$$

D
$$\displaystyle\frac{9}{4}\%$$

Solution

In an adiabatic process, $$PV^{\gamma}=constant$$
$$\gamma=\dfrac{C_P}{C_V}=\dfrac{3}{2}$$
Thus $$P_0V_0^{\gamma}=P_0(1+\dfrac{2/3}{100})V^{\gamma}$$
$$\implies V=\dfrac{V_0}{(1+\dfrac{2/3}{100})^{1/ \gamma}}$$
$$=V_0(1+\dfrac{2/3}{100})^{-2/3}$$
$$\approx V_0(1-\dfrac{4/9}{100})$$
Thus, volume decreases by $$\dfrac{4}{9}$$%
Question 9
1 / -0

For process 1, $$\Delta U$$ is positive, for process 2, $$\Delta U$$ is zero and for process 3, $$\Delta U$$ is negative. Find the parameters indicating $$x$$ and $$y$$ axes.

A
Temperature and volume

B
Temperature and pressure

C
No. of moles and pressure

D
Volume and pressure

Solution

$$\Delta U=0 \Rightarrow \Delta T=0\Rightarrow$$ isothermal process
Similarly $$\Delta T >0$$ for 1 and $$<0$$ for 3.
Thus starting from same pressure(P) and volume(V) and reaching same final volume the values of pressure will be $$P_1>P_2>P_3$$
Clearly x, y axes are thus volume and pressure axes respectively.
Question 10
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, find the pressure at a POINT on this straight line at which temperature is maximum

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

B
$$\displaystyle\dfrac{5}{4}P_o$$

C
$$\displaystyle\dfrac{3}{2}P_o$$

D
$$\displaystyle\dfrac{7}{5}P_o$$

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 $$ \dfrac{3 {V}_{o}}{2} $$

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Thermodynamics Test - 34

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Thermodynamics Test - 34

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Question 1

greater than the final pressure of $$B$$

equal to the final pressure of $$B$$

less than the final pressure of $$B$$

twice the final pressure of $$B$$

Solution

Question 2

$$C$$ and $$D$$, respectively

$$D$$ and $$C$$, respectively

$$A$$ and $$B$$, respectively

$$B$$ and $$A$$, respectively

Solution

Question 3

$$199800$$

$$0$$

$$200000$$

$$200200$$

Solution

Question 4

$$199800$$

$$0$$

$$200000$$

$$200200$$

Solution

Question 5

$$300\:K$$

$$250\:K$$

$$200\:K$$

$$100\:K$$

Solution

Question 6

$$2 \gamma$$

$$0$$

$$-\gamma$$

$$-1$$

Solution

Question 7

decrease

increase

remain same

none of these

Solution

Question 8

$$\displaystyle\frac{4}{9}\%$$

$$\displaystyle\frac{2}{3}\%$$

$$4\%$$

$$\displaystyle\frac{9}{4}\%$$

Solution

Question 9

Temperature and volume

Temperature and pressure

No. of moles and pressure

Volume and pressure

Solution

Question 10

$$\displaystyle\dfrac{4P_o}{3}$$

$$\displaystyle\dfrac{5}{4}P_o$$

$$\displaystyle\dfrac{3}{2}P_o$$

$$\displaystyle\dfrac{7}{5}P_o$$

Solution

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