Thermodynamics Test

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Thermodynamics Test - 50

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Question 1
1 / -0

One mole of an ideal gas is taken from state $$P$$ to state $$Q$$ by three different processes (i) $$PRQ$$, (ii) $$PSQ$$ and (iii) $$PTQ$$ as shown in the $$p - V$$ diagram. The heat absorbed by the gas is

A
The least in process (ii)

B
Greater in process (ii) than in process (i)

C
The same in the processes (i) and (ii)

D
Less in process (iii) than in process (ii)

Solution

Heat absorbed by gas in three processes can be written as
$$\triangle Q_{PRQ} = \triangle U + W_{PRQ}$$
$$\triangle Q_{PSQ} = \triangle U$$
$$\triangle Q_{PTQ} = \triangle U + W_{PTQ}$$
The change in internal energy in all the three cases is same and $$\triangle W_{PRQ} > 0, \triangle W_{PTQ} < 0$$
Hence, $$\triangle Q_{PRQ} > \triangle Q_{PSQ} > \triangle Q_{PTQ}$$.
Question 2
1 / -0

In a given process on an ideal gas $$ dW=0$$ and $$dQ<0,$$ then the force of gas

A
the temperature will decrease

B
the volume will increase

C
the pressure will remain constant

D
the temperature will increase

Solution

From first law of thermodynamics,
$$ dQ=dU+dW$$
we have $$dQ=dU (as\,dW=0)$$
But $$dQ<0$$
$$dU<0$$
$$n\,C_v \Delta T<0$$
or $$ \Delta T <0$$
Hence,the temperature will decrease
Question 3
1 / -0

For adiabatic expansion of perfect monoatomic gas , when voulme increases by $$ 24\%$$ ,what is the percentage decrease in pressure ?

A
$$ 24\% $$

B
$$ 30\% $$

C
$$ 48\% $$

D
$$ 71\% $$

Solution

In adiabatic process $$ p_2 V_2^{\gamma}=p_1 V_1^{\gamma}$$
$$p_2=p_1 (\dfrac{V_1}{V_2})^{\gamma}$$
$$p_2=(\dfrac{100}{124})^{5/3} p_1$$
$$ p_2=0.6985 p_1$$
$$\%$$ decrease in pressure
$$=\dfrac{p_1-p_2}{p_1}\times 100 \%$$
$$=\dfrac{p_1-0.6985 p_1}{p_1}\times 100 \%$$
$$=\dfrac{0.3015 p_1}{p_1} \times 100\%$$
$$=30.15 \%=30\%$$
Question 4
1 / -0

The p-V diagram of a gas system undergoing cyclic process is shown here. The work done during isobaric compression is

A
100 J

B
200 J

C
600 J

D
500 J

E
400 J

Solution

From P-V diagram, DA is the isobaric compression. Here pressure is constant i.e. $$p=2\times 10^{2} Nm^{-2}$$
Now, work done for DA is $$W=pdV=p(V_D-V_A) = (2\times 10^2)(3-1)=400 \, J$$
Question 5
1 / -0

A gas is compressed at a constant pressure of $$50\, N/m^2$$ from a volume of $$10m^3$$ to a volume of $$4 m^3$$. Energy of 100 J is then added to the gas by heating. Its internal energy is

A
Decreased by 200 J

B
Increased by 200 J

C
Decreased by 400 J

D
Increased by 400 J

Solution

From first law of the thermodynamics
$$\Delta Q=\Delta U + \Delta W = \Delta U + p\Delta V$$
$$\Rightarrow 100 = \Delta U+50 \times (4-10)$$
$$\Rightarrow \Delta U = 400\, J$$
Question 6
1 / -0

The $$p-V$$ plots for two gases during adiabatic process as shown in figure plots $$1$$ and $$2$$ should correspond respectively to.

A
$$He$$ and $$O_2$$

B
$$O_2$$ and $$He$$

C
$$He$$ and $$Ar$$

D
$$O_2$$ and $$N_2$$

Solution

Slope of adiabatic process of p-V diagram
$$\left(\displaystyle\frac{dp}{dv}\right)=\gamma \frac{p}{V}$$ i.e., slope $$\alpha\gamma$$
From graph$$(slope)_2 > (slope)_2$$
$$\Rightarrow \gamma_2 > \gamma_1$$
$$\gamma$$ for monoatomic gas (Ar, He) is greater than $$\gamma$$ for diatomic gas $$(O_2, N_2)$$.
Question 7
1 / -0

Two identical vessels $$A$$ and $$B$$ contain equal amount of ideal monoatomic gas. The piston of $$A$$ is fixed but that of $$B$$ is free. Same amount of heat is absorbed by $$A$$ and $$B$$. If $$B's$$ internal energy increases by $$100\ J$$ the change in internal energy of $$A$$ is

A
$$100\ J$$

B
$$\dfrac {500}{3}J$$

C
$$250\ J$$

D
None of these

Solution

From $$1st$$ Law of thermodynamics, we have $$dQ=dU+dW=dU+PdV\rightarrow(1)$$
For case $$A:dV=0$$ So, $$dW=0$$ and hence
$$\Rightarrow dQ_A=dU_A\rightarrow(2)$$
For case $$B:$$ It is free expansion so, $$\triangle W$$ has to be zero and that's why
$$\Rightarrow dQ_B=dU_B=100J\rightarrow (3)$$
But $$ dQ_A= dQ_B=100J$$ So, using $$(2)$$ we get,
$$\Rightarrow dU_A=100J,$$ Change in internal energy of $$A.$$
Hence, the answer is $$100J.$$
Question 8
1 / -0

A gas at pressure $$p$$ has volume $$V$$. It is adiabatically compressed to volume $$V/32$$. If $$(32)^{1.4} = 128$$, what is the current pressure of the gas?

A
$$128p$$

B
$$63p$$

C
$$32p$$

D
$$16p$$

Solution

$$P{ V }^{ \gamma }=Constant\\ { P }_{ 1 }{ V }_{ 1 }^{ \gamma }={ P }_{ 2 }{ V }_{ 2 }^{ \gamma }\\ { P }_{ 2 }={ P }_{ 1 }\times ({ \cfrac { { V }_{ 1 } }{ { V }_{ 2 } } })^{ \gamma }\\ =P\times ({ \cfrac { V }{ \cfrac { V }{ 32 } } })^{ 1.4 }\quad [\gamma =1.4]\\ =P\times ({ 32) }^{ 1.4 }\\ { P }_{ 2 }=128P$$
Question 9
1 / -0

For an ideal gas with initial pressure and volume $$P_i$$ and $$V_i$$ respectively, a reversible isothermal expansion happens, when its volume becomes $$V_0$$. Then it is compressed to its original volume $$V_i$$ by a reversible adiabatic process. If the final pressure is $$P_t$$ then which of the following statements is true?

A
$$P_f = P_i$$

B
$$P_f > P_i$$

C
$$P_f < P_i$$

D
$$\displaystyle\frac{P_f}{V_o}=\frac{P_i}{V_i}$$

Solution

It is clear from the image that $$P_f>P_i$$ .
Question 10
1 / -0

The adiabatic Bulk modulus of a diatomic gas at atmospheric pressure is

A
$$0\ Nm^{-2}$$

B
$$1\ Nm^{-2}$$

C
$$1.4\times 10^{4}\ Nm^{-2}$$

D
$$1.4\times 10^{5}\ Nm^{-2}$$

Solution

The correct answer is option(D).
We have the bulk modulus formula as:
$$\text{Bulk Modulus}=\dfrac {\text{Pressure}}{\text{Strain}}=\dfrac {p}{(V_0-V_n)/V_0}$$
For an adiabatic process, the bulk modulus is given by
$$k=-\dfrac {V\Delta p}{\Delta V}=\gamma p$$
adiabatic bulk modulus $$=\gamma p$$
At NTP, $$p=1.013\times 10^5N/m^2$$ and $$\gamma=1.4$$
Hence Bulk modulus $$=1.013\times 10^5\times 1.4\approx 1.4\times 10^5N/m^2$$

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Thermodynamics Test - 50

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Thermodynamics Test - 50

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SHARING IS CARING

Question 1

The least in process (ii)

Greater in process (ii) than in process (i)

The same in the processes (i) and (ii)

Less in process (iii) than in process (ii)

Solution

Question 2

the temperature will decrease

the volume will increase

the pressure will remain constant

the temperature will increase

Solution

Question 3

$$ 24\% $$

$$ 30\% $$

$$ 48\% $$

$$ 71\% $$

Solution

Question 4

100 J

200 J

600 J

500 J

400 J

Solution

Question 5

Decreased by 200 J

Increased by 200 J

Decreased by 400 J

Increased by 400 J

Solution

Question 6

$$He$$ and $$O_2$$

$$O_2$$ and $$He$$

$$He$$ and $$Ar$$

$$O_2$$ and $$N_2$$

Solution

Question 7

$$100\ J$$

$$\dfrac {500}{3}J$$

$$250\ J$$

None of these

Solution

Question 8

$$128p$$

$$63p$$

$$32p$$

$$16p$$

Solution

Question 9

$$P_f = P_i$$

$$P_f > P_i$$

$$P_f < P_i$$

$$\displaystyle\frac{P_f}{V_o}=\frac{P_i}{V_i}$$

Solution

Question 10

$$0\ Nm^{-2}$$

$$1\ Nm^{-2}$$

$$1.4\times 10^{4}\ Nm^{-2}$$

$$1.4\times 10^{5}\ Nm^{-2}$$

Solution

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