Thermodynamics Test

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Thermodynamics Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following statements is incorrect?

A
It is not possible to compress a gas at a temperature below $$T_c$$

B
At a temperature below $$T_c$$, the molecules are close enough for the attractive forces to act, and condensation occur

C
No condensation takes place above $$T_c$$

D
The kinetic energy of the gas molecules is higher above $$T_c$$, and the attraction between them decreases

Solution

Critical temperature of the substance is the temperature at a substance's critical point.
Above $$T_c$$ we cannot liquify gas keep pressure constant although on $$\uparrow$$ pressure we can liquify.
Question 2
1 / -0

In a cylinder, $$2.0 \ moles$$ of an ideal monatomic gas initially at $$1.0 \times 10^6 \ Pa$$ and $$300 K$$ expands until its volume doubles. The work done if the expansion is isobaric is

A
$$4980\ J$$

B
$$2800\ J$$

C
$$2300\ J$$

D
$$4630 \ J$$

Solution

Work done in isobaric process=$$\Delta W=P\Delta V=nR\Delta T$$
Here, n=no. of moles=2
At constant pressure, volume doubles$$ \Rightarrow$$ temperature doubles $$\Rightarrow \Delta T=T_o=300K$$
Thus,$$ \Delta W=2 \times 8.3\times 300 \ J=4980 \ J$$
Question 3
1 / -0

Directions For Questions

Piston-cylinder device initially contains $$0.5 m^3$$ of nitrogen gas at $$400 kPa$$ and $$27^o$$C. An electric heater within the device is at turned on and is allowed to pass a current of $$2 A$$ for $$5min$$ from a $$120 V$$ source. Nitrogen expands at constant pressure and a heat loss of $$2800 J$$ occurs during the process.$$R = 8.314 kJ/kmol-K$$

...view full instructions

Electric work done on the nitrogen gas is

A
$$72\ kJ$$

B
$$36\ kJ$$

C
$$18\ kJ$$

D
$$9\ kJ$$

Solution

Electric work=energy supplied via the appliance= $$ V \times I \times t$$
=$$120\times 2\times 300J=72 \ kJ$$
Question 4
1 / -0

Two moles of an ideal gas at temperature $$T_o = 300 K$$ was cooled isochorically so that the pressure was reduced to half. Then, in an isobaric process, the gas expanded till its temperature got back to the initial value. The total amount of heat absorbed by the gas in the process is:

A
2490 J

B
2680 J

C
240 J

D
290 J

Solution

Let the initial thermodynamic coordinates be ($$P_0,V_0,T_0$$),followed by ($$P_1,V_1,T_1$$) and finally ($$P_2,V_2,T_2$$)
Given $$T_0=300K,P_1=\dfrac{P_0}{2}$$
First process being isochoric, $$V_1=V_0$$
Thus, $$T_1=\dfrac{T_0}{2}=150K$$
Also, $$T_2=T_0=2T_1=300K$$
So, heat released in isochoric cooling=$$\Delta Q_1=nC_v\Delta T$$
and heat absorbed in isobaric heating=$$\Delta Q_2=nC_p\Delta T$$
Net heat absorbed=$$\Delta Q_2-\Delta Q_1=n(C_p-C_v)\Delta T=nR\Delta T=2\times 8.3\times 150 \ J=2490 \ J$$
Question 5
1 / -0

In an adiabatic process, the pressure is increased by$$\frac{2}{3}$$%. If $$\gamma =\frac{3}{2}$$, then the volume decreases by nearly:

A
$$\dfrac{4}{9}$$%

B
$$\dfrac{2}{3}$$%

C
$$1$$ %

D
$$\dfrac{9}{4}$$%

Solution
Question 6
1 / -0

As a result of isobaric heating by $$\Delta T = 72 K$$, one mole of certain ideal gas obtains an amount of heat $$Q = 1.60 kJ$$. The value of $$\gamma$$ is

A
$$1.60$$

B
$$1.33$$

C
$$1$$

D
$$0.66$$

Solution

For isobaric process, heat absorbed $$\Delta Q=nC_p\Delta T$$
where, n = no. of moles = 1,
$$\Rightarrow C_p=\dfrac{1600}{72}
\Rightarrow \gamma=\dfrac{C_p}{C_p-R}=1.6$$
Question 7
1 / -0

An ideal gas initially at $$30 K$$ undergoes an isobaric expansion at $$2.50 \ kPa$$. If the volume increases from $$1.00 m^3$$ to $$3.00 m^3$$ and $$12.5 \ kJ$$ is transferred to the gas by heat,what are its final temperature?

A
$$900 K$$

B
$$400 K$$

C
$$500 K$$

D
$$200 K$$

Solution

Applying ideal gas equation for initial state ($$P_1,V_1,T_1$$)and final state ($$P_2,V_2,T_2$$) resp. we get,
$$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$$

Since isobaric, $$ P_1=P_2$$
$$\Rightarrow T_2=\dfrac{T_1V_2}{V_1} =3T_1=900K$$
Question 8
1 / -0

In the following pressure-volume diagram, the isochoric, isothermal and isobaric parts, respectively, are

A
BA, AD, DC

B
DC, CB, BA

C
AB, BC, CD

D
CD, DA, AB

Solution

We know that $$PV = nRT$$
If volume V is constant, process is called isochoric. So, CD is isochoric.
If pressure P is constant , process is called isobaric. So, AB is isobaric.
If temperature T is constant, process is called isothermal.
If T is constant than PV = constant. So, graph will be rectangular hyperbola.
So, BC and AD are isothermal process.
Question 9
1 / -0

Figure shows four PV diagrams. Which of these curves represent isothermal and adiabatic processes respectively?

A
C and D

B
A and C

C
A and B

D
B and D

Solution

During an adiabatic process slope of PV curve is: $$\dfrac{dP}{dV}=-\gamma \dfrac{P}{V}$$
During an isothermal process slope of PV curve: $$\dfrac{dP}{dV}= -\dfrac{P}{V}$$
Both slopes are negative hence A and B are not possible since the slopes for curves A and B are positive.
Slope of adaibatic curve is more than the slope of isothermal curve, so C is isothermal and D is adiabatic.
Question 10
1 / -0

A gas is at $$1\:atm$$ pressure with a volume $$800\:cm^3$$. When $$100\:J$$ of heat is supplied to the gas, it expands to $$1\:L$$ at constant pressure. The change in its internal energy is

A
$$80\:J$$

B
$$-80\:J$$

C
$$20\:J$$

D
$$-20\:J$$

Solution

The work done on the gas at constant pressure is,
$$ W = P \Delta V $$
Thus, W = $$ 1 \times (1 - 0.8) $$
$$W = 0.2 \ atm-L = 20 J$$
Now, heat supplied, $$Q = 100 J$$
According to the first law of thermodynamics,
Q = U + W
Thus, $$U = 100 - 20 = 80 J$$

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Thermodynamics Test - 33

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Thermodynamics Test - 33

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Question 1

It is not possible to compress a gas at a temperature below $$T_c$$

At a temperature below $$T_c$$, the molecules are close enough for the attractive forces to act, and condensation occur

No condensation takes place above $$T_c$$

The kinetic energy of the gas molecules is higher above $$T_c$$, and the attraction between them decreases

Solution

Question 2

$$4980\ J$$

$$2800\ J$$

$$2300\ J$$

$$4630 \ J$$

Solution

Question 3

$$72\ kJ$$

$$36\ kJ$$

$$18\ kJ$$

$$9\ kJ$$

Solution

Question 4

2490 J

2680 J

240 J

290 J

Solution

Question 5

$$\dfrac{4}{9}$$%

$$\dfrac{2}{3}$$%

$$1$$ %

$$\dfrac{9}{4}$$%

Solution

Question 6

$$1.60$$

$$1.33$$

$$1$$

$$0.66$$

Solution

Question 7

$$900 K$$

$$400 K$$

$$500 K$$

$$200 K$$

Solution

Question 8

BA, AD, DC

DC, CB, BA

AB, BC, CD

CD, DA, AB

Solution

Question 9

C and D

A and C

A and B

B and D

Solution

Question 10

$$80\:J$$

$$-80\:J$$

$$20\:J$$

$$-20\:J$$

Solution

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