Thermodynamics Test

Result

Thermodynamics Test - 44

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Figure below shows pressure (P) versus volume(V) graphs for a certain mass of a gas at two constant temperatures $$T_1$$ and $$T_2$$. Which of the inferences given below is correct?

A
$$T_1 = T_2$$

B
$$T_1 > T_2$$

C
$$T_1 < T_2$$

D
none of the above

Solution

For a given volume V, pressure at curve $$T_1$$ will be lower than that at curve $$T_2$$. Therefore, $$P_2$$ > $$P_1$$ and so $$T_2$$ > $$T_1$$.
Question 2
1 / -0

An ideal gas at $${27}^{o}C$$ is compared adiabatically to $$\dfrac{8}{27}$$ its original volume [$$T{V}^{\gamma-1}=$$constant] and $$\gamma=\dfrac{5}{3}$$, then the rise in temperature will be:

A
$${480}^{o}C$$

B
$${450}^{o}C$$

C
$${375}^{o}C$$

D
$${225}^{o}C$$

Solution

Applying the formula For adiabatic Process

$${T}_{1}{V}_{1}^{\gamma-1}={T}_{2}{V}_{2}^{\gamma-1}$$
$$\cfrac{{T}_{1}}{{T}_{2}}={ \left( \cfrac { { V }_{ 2 } }{ { V }_{ 1 } } \right) }^{ \gamma -1 }$$
$$ \implies \left( \cfrac { 27+273 }{ { T }_{ 2 } } \right) ={ \left( \cfrac { 8 }{ 27 } \right) }^{ \gamma -1 }={ \left( \cfrac { 8 }{ 27 } \right) }^{ 5/3-1 }$$
or, $$\cfrac{300}{{T}_{2}}={ \left( \cfrac { 2 }{ 3 } \right) }^{ \cfrac { 2 }{ 3 } \times 3 }=\cfrac { 4 }{ 9 } $$
$${T}_{2}=\cfrac{9\times 300}{4}={675}^{o}K$$
$${t}^{o}C=675-273={402}^{o}C$$
Rise in temperature $$=402-27={375}^{o}C$$
Question 3
1 / -0

Which of the following parameters does not characterise the thermodynamic state of matter?

A
Temperature

B
Pressure

C
Work

D
Volume

Solution

Thermodynamics state of matter is characterized by temperature, pressure and volume only, but not work.
Work done by a gas is defined as, $$W = -\int Pdv$$
The thermodynamic state of a system is defined by specifying values of a set of measurable properties sufficient to determine all other properties. The thermodynamic variables in case of a gas are pressure, temperature, and volume in addition to number of moles.
Question 4
1 / -0

A gas expands adiabatically at constant pressure, such that its temperature $$T \propto \frac{1}{\sqrt{V}}$$. The value of $$C_p / C_V$$ of the gas is:

A
1.30

B
1.50

C
1.67

D
2.00

Solution

Hint: Use adiabatic relation between pressure $$P$$ and volume $$V$$
$$PV^{\gamma}=$$ constant
Explanation:
Step 1: Formula used:
For adiabatic expansion, we have the formula
$$PV^{\gamma}=$$ constant ..............(i)

For ideal gas equation is,
$$PV=RT$$
$$\Rightarrow P=\dfrac{RT}{V}$$ ............(ii)
From Eqs. (i) and (ii), we obtain
$$\left(\dfrac{RT}{V}\right)V^{\gamma}=$$ constant
$$\Rightarrow TV^{\gamma-1}=$$constant ...........(iii)

Step 2: Given:
$$T\propto \dfrac{1}{\sqrt{V}}$$
as $$TV^{1/2}=$$ constant ...(iv)

Step 3: Conclusion:
Thus, using Eqs. (iii) and (iv) together, we get
$$\gamma - 1=\dfrac{1}{2}$$
or $$\gamma =\dfrac{3}{2}=1.5$$
$$\Rightarrow \dfrac{C_p}{C_v}= 1.5$$

Option B is correct
Question 5
1 / -0

The pressure $$p$$, volume $$V$$ and temperature $$T$$ for a certain gas are related by $$p=\dfrac{AT-BT^2}{V}$$, where $$A$$ and $$B$$ are constants. the work done by the gas when the temperature changes from $$T_1$$ to $$T_2$$ while the pressure remains constant, is given by:

A
$$A(T_2 - T_1) - B(T_2^2-T_1^2)$$

B
$$\displaystyle \frac{A(T_2-T_1)}{V_2-V_1}-\frac{B(T_2^2-T_1^2)}{V_2-V_1}$$

C
$$A(T_2-T_1)-\frac{B}{2}(T_2^2-T_1^2)$$

D
$$\displaystyle \frac{A(T_2-T^2_1)}{V_2-V_1}$$

Solution

Given, $$P=\dfrac{AT-BT^2}{V}$$
$$\Rightarrow PV=AT-BT^2$$
$$\Rightarrow P\Delta V=A\Delta T - 2BT\Delta T$$
On integrating, we get
Work $$=\int PdV=A\int _{T_1}^{T_2}dT-2B\int_{T_1}^{T_2}TdT$$
$$=A(T_2-T_1)- {B}[(T_2)^2-(T_1)^2]$$
Question 6
1 / -0

A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in the figure. Its volume is then reduced to the original value from $$E$$ to $$F$$ by an isobaric process. Calculate the total work done by the gas from $$D$$ to $$E$$ to $$F$$.

A
$$225\ J$$

B
$$450\ J$$

C
$$900\ J$$

D
$$600\ J$$

Solution

Total work done by the gas from $$D$$ to $$E$$ to $$F$$ = Area of $$\triangle DEF$$
Area of $$\triangle DEF = (1/2)DE\times EF$$
Where,
$$DF$$ = Change in pressure
$$= 600 N/m^2-300N/m^2$$
$$= 300 N/m^2$$
$$FE$$ = Change in volume
$$= 5.0 m^3-2.0m^3$$
$$= 3.0 m^3$$
Area of $$\triangle DEF = (1/2)\times 300 \times 3 = 450 J$$
Therefore, the total work done by the gas from $$D$$ to $$E$$ to $$F$$ is $$450 J$$.
Question 7
1 / -0

In an adiabatic change, the pressure and temperature of a monoatomic gas are related as $$P\propto {T}^{C}$$, where $$C$$ equals

A
$$\cfrac{2}{5}$$

B
$$\cfrac{5}{2}$$

C
$$\cfrac{3}{5}$$

D
$$\cfrac{5}{3}$$

Solution

In an adiabatic process
$$P^{\gamma -1}T^{\gamma}=constant$$
Hence, $${P}^{\gamma-1}\propto {T}^{\gamma}$$ where $$\gamma=\cfrac{5}{3}$$ for monoatomic gases
for monoatomic gas $$\therefore$$ $$P\propto {T}^{\gamma/(\gamma-1)}$$
$$\therefore$$ $$C=\cfrac{\gamma}{\gamma-1}=\cfrac{5/3}{5/3-1}=\cfrac{5/3}{2/3}=\cfrac{5}{2}$$
Question 8
1 / -0

In an adiabatic change, the pressure and temperature of a monoatomic gas are related with relation as $$P\propto {T}^{C}$$, where $$C$$ is equal to :

A
$$\cfrac{5}{4}$$

B
$$\cfrac{5}{3}$$

C
$$\cfrac{5}{2}$$

D
$$\cfrac{3}{5}$$

Solution

$$Pk{T}^{C}$$ $$\Rightarrow$$ $$P{T}^{-C}=k$$
For adiabatic change
$$P{V}^{\gamma}=$$ constant [where $$\gamma$$ is $$\cfrac{{C}_{p}}{{C}_{v}}$$]
or $$P{T}^{\gamma/1-\gamma}=$$ constant
Comparing it with the given relation
$$\cfrac{\gamma}{1-\gamma}=-C$$
$$\cfrac{\gamma}{\gamma-1}=C$$
For monoatomic gas, $$\gamma=\cfrac{5}{3}$$
$$\cfrac{5/3}{\cfrac{5}{3}-1}=C$$ $$\Rightarrow$$ $$C=\cfrac{5}{3}\times \cfrac{3}{2}=\cfrac{5}{2}$$
Question 9
1 / -0

An ideal gas is compressed isothermally until its pressure is doubled and then allowed to expand adiabatically to regain its original volume ($$\gamma=1.4$$ and $${2}^{-1.4}=0.38)$$. The ratio of the final to initial pressure is

A
$$0.76:1$$

B
$$1:1$$

C
$$0.66:1$$

D
$$0.86:1$$

Solution

Let initial pressure, volume and temperature be $$P$$, $$V$$ and $$T$$ respectively.
$$\therefore$$ $$P_2 = 2P$$

For isothermal process : $$P_2V_2 = P_1V_1$$
$$(2P) V_2 = PV$$ $$\implies V_2 = \dfrac{V}{2}$$

For adiabatic process : $$P_2 V_2^{\gamma} = P_3V_3^{\gamma}$$ (Given $$\gamma = 1.4$$)
$$\Rightarrow 2P \bigg(\dfrac{V}{2}\bigg)^{1.4} = P_3V^{1.4}$$

$$\Rightarrow$$ $$2 \bigg(\dfrac{1}{2}\bigg)^{1.4} = \dfrac{P_3}{P}$$

$$\Rightarrow 2 \times 0.38 = \dfrac{P_3}{P}$$

$$\Rightarrow \dfrac{P_3}{P} = 0.76$$

Thus, $$P_3:P = 0.76 : 1$$
Question 10
1 / -0

Blowing air with open mouth is an example of :

A
isobaric process

B
isochoric process

C
isothermal process

D
adiabatic process

Solution

When air is blown with open mouth, its pressure is same as that of atmospheric pressure since it is in direct contact with atmosphere. Hence it is an isobaric process.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Thermodynamics Test - 44

Result

Thermodynamics Test - 44

Score

-

Rank

-

SHARING IS CARING

Question 1

$$T_1 = T_2$$

$$T_1 > T_2$$

$$T_1 < T_2$$

none of the above

Solution

Question 2

$${480}^{o}C$$

$${450}^{o}C$$

$${375}^{o}C$$

$${225}^{o}C$$

Solution

Question 3

Temperature

Pressure

Work

Volume

Solution

Question 4

1.30

1.50

1.67

2.00

Solution

Question 5

$$A(T_2 - T_1) - B(T_2^2-T_1^2)$$

$$\displaystyle \frac{A(T_2-T_1)}{V_2-V_1}-\frac{B(T_2^2-T_1^2)}{V_2-V_1}$$

$$A(T_2-T_1)-\frac{B}{2}(T_2^2-T_1^2)$$

$$\displaystyle \frac{A(T_2-T^2_1)}{V_2-V_1}$$

Solution

Question 6

$$225\ J$$

$$450\ J$$

$$900\ J$$

$$600\ J$$

Solution

Question 7

$$\cfrac{2}{5}$$

$$\cfrac{5}{2}$$

$$\cfrac{3}{5}$$

$$\cfrac{5}{3}$$

Solution

Question 8

$$\cfrac{5}{4}$$

$$\cfrac{5}{3}$$

$$\cfrac{5}{2}$$

$$\cfrac{3}{5}$$

Solution

Question 9

$$0.76:1$$

$$1:1$$

$$0.66:1$$

$$0.86:1$$

Solution

Question 10

isobaric process

isochoric process

isothermal process

adiabatic process

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.