Thermodynamics Test

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Thermodynamics Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

The $$P-T$$ diagram for an ideal gas is shown in Fig., where $$AC$$ is an adiabatic process, find the corresponding $$P-V$$ diagram.

A

B

C

D

Solution

process AB is isothermal DV = constant will show a hyperbol curve process AC adiabatic it will show another curve,process BC is isochonic it will show a strength line in P-V digram
Question 2
1 / -0

$$0.5$$ mole of oxygen and $$0.5$$ mole of nitrogen, each having volume $$V$$ and temperature $$T$$ are mixed isothermally to have the total volume $$2V$$. The maximum work done is:

A
$$R T \log _ { e } 2$$ joule

B
$$\dfrac { R T } { 2 } \log _ { e } 2$$ joule

C
$$R T \left( \log _ { e } 4 \right)$$ joule

D
zero
Question 3
1 / -0

State the equation corresponding to 8 g of $$O_{2}$$ is then

A
Pv = 8 RT

B
PV = RT /4

C
PV = RT

D
PV = RT /2

Solution

Given

$$ 16\,g\,{{O}_{2}}=1\,mole\,of\,{{O}_{2}} $$

$$ 8\,g\,{{O}_{2}}=0.5\,mole\,of\,{{O}_{2}} $$

Number of mole in $$8\,g\,{{O}_{2}}$$ = $$0.5\,mole=\,\dfrac{1}{2}\,mole$$

From ideal gas equation,

$$ PV=nRT $$

$$ PV=\dfrac{RT}{2} $$

Hence, $$PV=\dfrac{RT}{2}$$is the relation.
Question 4
1 / -0

A certain quantity of ideal gas takes up $$6$$ $$\mathrm { J }$$ fheat in process $$AB$$ and $$360J$$ in process AC. The adiabatic constant for the gas is:

A
$$\frac { 7 } { 5 }$$

B
$$\frac { 8 } { 6 }$$

C
$$\frac { 5 } { 3 }$$

D
$$\frac { 6 } { 4 }$$
Question 5
1 / -0

The pressure and density of a diatomic gas ($$\gamma = 7/5$$) change from $$(P_{1}, d_{1}) to (P_{2}, d_{2})$$ adiabatically . If $$d_{2}/d_{1}=32$$, then what is the value of $$P_{2}/P_{1}$$?

A
$$\dfrac{P}{P'} = \dfrac{1}{128}$$

B
$$\dfrac{V'}{V} = \dfrac{1}{128}$$

C
$$\dfrac{P}{P'} = 128$$

D
$$\dfrac{V'}{V} = 32$$

Solution

Given : The pressure and density of a diatomic gas (γ=7/5γ=7/5) change from (P1,d1)to(P2,d2)(P1,d1)to(P2,d2) adiabatically . If d2/d1=32d2/d1=32, then what is the value of P2/P1P2/P1?

Solution :
We know that for adiabatic process $$PV^{\gamma}= constant $$
As $$ Density =\frac{mass}{volume}$$

Then $$\frac{V_1}{V_2}=\frac{d_2}{d_1}$$

Hence we can say that $$\frac{P_1}{P_2}=(\frac{d_2}{d_1})^{\gamma}$$
$$\frac{P_1}{P_2}=(\frac{1}{32})^{\frac{7}{5}}$$
$$\frac{P_2}{P_1}=128$$

The Correct Opt=C
Question 6
1 / -0

During the adiabatic change of ideal gas, The relation between the pressure and the density will be -

A
$$\frac{P_{1}}{P_{2}}=\left ( \frac{d_{2}}{d_{1}} \right )^{1/\gamma }$$

B
$$p_{1}d^{\gamma }_{1}=p_{2}d_{2}\gamma $$

C
$$p_{1}d^{-\gamma }_{1}=p_{2}d_{2}^{-\gamma }$$

D
$$\frac{P_{1}}{P_{2}}=\left ( \frac{d_{2}}{d_{1}} \right )^{\gamma }$$

Solution

For an adiabatic process, $${{P}_{1}}V_{1}^{\gamma }={{P}_{2}}V_{2}^{\gamma }=cons\tan t$$

Since density,

$$ d=\dfrac{m}{V} $$

$$ V\propto \dfrac{1}{d} $$

$$ {{P}_{1}}{{\left( \dfrac{1}{{{d}_{1}}} \right)}^{\gamma }}={{P}_{2}}{{\left( \dfrac{1}{{{d}_{2}}} \right)}^{\gamma }} $$

$$ {{P}_{1}}d_{1}^{-\gamma }={{P}_{2}}d_{2}^{-\gamma } $$
Question 7
1 / -0

The internal energy change in a system that has absorbed $$2\ Kcal$$ of heat and done $$500\ J$$ of work is

A
$$8900\ J$$

B
$$6400\ J$$

C
$$5400\ J$$

D
$$7900\ J$$

Solution

Given: Heat absorbed = 2 Kcal
work done = 500 J
Solution: As we know, according to the law of thermodynamics
$$\bigtriangleup U=Q-W$$ (1)
where, U is the change in the internal energy,
Q denotes the heat given to the system and
W denotes the work done by the system.
Also, $$1 cal = 4.2 J$$
So, $$2 Kcal = 2\times 1000\times 4.2$$
Substituting the values in equation (1) we get,
$$\bigtriangleup U=2\times 1000\times 4.2-500$$
$$\bigtriangleup U=7900J$$
Hence, the correct option is (D).
Question 8
1 / -0

One mole gas is first cooled from 300 K to 150 K at constant volume and then heated from 150 K to 300 K at constant pressure. The net heat absorbed by the gas is

A
zero

B
150 R

C
300 R

D
450 R

Solution

Case:1
At constant volume,
Heat absorbed,$$\Delta Q_1=n C_v\Delta T$$
$$\Rightarrow \Delta Q_1= 1×C_v×(150-300) = -150C_v$$
Case:2
At constant pressure,
Heat absorbed,$$\Delta Q_2=nC_p \Delta T$$
$$\Rightarrow \Delta Q_2=1×C_p×(300-150)=150C_p$$
Therefore, net heat absorbed = $$\Delta Q_1+\Delta Q_2$$
$$\Rightarrow Net heat absorbed = 150(C_p-C_v)=150R$$

Hence, option B is correct.
Question 9
1 / -0

A constant volume gas thermometer shows pressure reading of 50 cm and 90 cm of mercury at 0$$^{0}C$$ and 100$$^{0}C$$ respectively. When the pressure reading is 60 cm of mercury, the temperature is:-

A
25$$^{0}C$$

B
40$$^{0}C$$

C
$$^{0}C$$

D
$$30^{0}C$$

Solution

Given,

Temperature $${{T}_{1}}={{0}^{o}}C$$ and height $${{h}_{1}}=50\,cm$$

Temperature$${{T}_{2}}=100{{\,}^{o}}C$$ and height $${{h}_{2}}=90\,cm$$

$$ \dfrac{T-{{T}_{1}}}{h-{{h}_{1}}}=\dfrac{{{T}_{2}}-{{T}_{1}}}{{{h}_{2}}-{{h}_{1}}} $$

$$ \dfrac{T-0}{60-50}=\dfrac{100-0}{90-50} $$

$$ T=25{{\,}^{o}}C $$

Hence, at $$60 cm $$, temperature is $$25{{\,}^{o}}C$$
Question 10
1 / -0

Dry air at $$15C$$ and $$10$$ $$atm$$ is suddenly released in atmospheric pressure. Find the final temperature of the air. (for air, $$\cfrac{Cp}{Cv}=1.41$$)

A
$$108K$$

B
$$148K$$

C
$$198K$$

D
$$298K$$

Solution