Thermodynamics Test

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Thermodynamics Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following represents an isothermal expansion of an ideal gas ?

A

B

C

D

Solution

Isothermal $$PV$$ is a rectangular hyperbola. Hence, option D.
Question 2
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For adiabatic expansion of a monoatomic perfect gas, the volume increases by $$2.4\%$$. What is the percentage decreases in pressure ?

A
$$2.4\%$$

B
$$4.0\%$$

C
$$4.8\%$$

D
$$7.1\%$$

Solution

$$\textbf{Hint:}$$ pressure is define as force per unit area
$$\textbf{Step 1:}$$ For an adiabatic process
$$\mathrm{PV}^{\mathrm{r}}=$$ constant
Differentiating on both sides: $$\mathrm{V}^{\mathrm{r}} \delta \mathrm{P}+\gamma \mathrm{PV}^{\gamma-1} \delta \mathrm{V}=0$$
$$\textbf{Step 2:}$$ dividing by
$$\mathrm{PV}^{\gamma}=$$ constant$$\dfrac{\delta \mathrm{P}}{\mathrm{P}}+\gamma \dfrac{\delta \mathrm{V}}{\mathrm{V}}=0$$
$$\textbf{Step 3:} $$Finding increase in percentage
Percentage increase in volume is $$2.4 \%$$
i.e. $$\dfrac{\delta \mathrm{V}}{\mathrm{V}}=\dfrac{2.4}{100}$$ and $$\gamma=\dfrac{5}{3}$$
i.e. $$\dfrac{\delta \mathrm{P}}{\mathrm{P}}=-\gamma \dfrac{\delta \mathrm{V}}{\mathrm{V}}=-\dfrac{5}{3} \times \dfrac{2.4}{100}=-\dfrac{4}{100}$$
.

Hence decrease in pressure is $$4\%$$
Question 3
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The pressure and volume of gas(air) are $$P$$ and $$V$$ respectively. If it is compressed suddenly to $$\dfrac{1}{32}$$ of its initial volume then its final pressure will be

A
$$\dfrac{P}{128}$$

B
$$\dfrac{P}{32}$$

C
$$128P$$

D
$$32P$$

Solution

Sudden compression is adiabatic as there is no scope for heat exchange in such short notice.
Thus $$PV^\gamma$$ =constant ,where $$\gamma=7/5$$
Hence, $$P_iV_i^{7/5}=P_fV_f^{7/5}$$
$$\Rightarrow PV^{7/5}= P_f(\dfrac{V}{32})^{7/5}\Rightarrow P_f=128P$$
Hence final pressure is $$128P$$.
Question 4
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$$1 m^3$$ of gas is compressed suddenly at atmospheric pressure and temperature $$27^oC$$ such that its temperature becomes $$627^oC$$. the final pressure of the gas (if $$\gamma=1.5$$) will be

A
$$2.7 \times 10^5 Nm^{-2}$$

B
$$7.2 \times 10^5 Nm^{-2}$$

C
$$27 \times 10^5Nm^{-2}$$

D
$$2.7 \times 10^6 Nm^{-2}$$
Solution
For an adiabatic process, the equation of state is given by:
$$T^{\gamma}P^{1 - \gamma} = constant$$
This implies,
$$T_{1}^{\gamma}P_{1}^{1-\gamma} = T_{2}^{\gamma}P_{2}^{1-\gamma}$$
Given:
$$T_{1} = 27^{\circ}C = 300K$$
$$T_{2} = 627^{\circ}C = 900K$$
$$P_{1} = 1.013 \times 10^{5} Pa$$
$$\gamma = 1.5$$
Substituting this into the equation of state:
We get $$P_{2} = 27 \times 10^{5} Pa$$
Question 5
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Isobaric bulk modules of elasticity is

A
$$\infty$$

B
$$0$$

C
$$P$$

D
$$\dfrac{C_p}{C_v}$$

Solution

Bulk modulus is defined as the ratio of the infinitesimal pressure increase to the resulting relative decrease of the volume. or $$K=-V\displaystyle\dfrac{dP}{dV}$$. But in a isobaric process pressure remains unchanged.
Thus bulk modulus K is zero.
Question 6
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The ratio of work done by an ideal diatomic gas to the heat supplied by the gas in an isobaric process is

A
$$\dfrac{5}{7}$$

B
$$\dfrac{3}{5}$$

C
$$\dfrac{2}{7}$$

D
$$\dfrac{5}{3}$$

Solution
Question 7
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A reversible engine takes heat from a reservoir at $$527 ^o$$ C and gives out heat to a sink at $$127^o$$ C. How many Joules/s must it take from the reservoir to perform useful mechanical work at the rate of $$750 \ W$$?

A
$$257.14$$

B
$$1500$$

C
$$457.14$$

D
$$557.14$$

Solution

Efficiency of a reversible heat engine is given by
$$\eta =1-\dfrac {T_c}{T_h}=\dfrac {W}{Q_h}$$
$$\implies 1-\dfrac {400}{800}=\dfrac {750}{Q_h}$$
$$\implies 0.5=\dfrac {750}{Q_h}$$
$$\therefore Q_h=\dfrac {750}{0.5}=1500J/s$$
Question 8
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A tyre pumped to a pressure $$3.375\ atm$$ at $$27^oC$$ suddenly bursts. What is the final temperature ($$\gamma = 1.5$$) ?

A
$$27^oC$$

B
$$-27^oC$$

C
$$0^oC$$

D
$$-73^oC$$

Solution

The given process is adiabatic as no heat transfer is involved. Thus we have
$$\displaystyle { T }_{1 }^{ \gamma }{ P}_{1 }^{1- \gamma } = { T }_{2 }^{ \gamma }{ P}_{2 }^{1- \gamma }$$
or $$\displaystyle \left(\dfrac { T_{ 1 } }{ T_{ 2 }}\right)^{\gamma}=\left(\dfrac{ P_{ 1 } }{ P_{ 2 } }\right)^{ \gamma -1 }$$
$$=\left(\displaystyle\dfrac { 300 }{ T_{ 2 } }\right)^{3/2 } =\left(\displaystyle\dfrac { 3.375 }{ 1 }\right)^{3/2-1} $$
or $$\displaystyle T_2 =(\dfrac{300}{3.375})^{1/3} = 200K = -73^o C$$
Question 9
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A monoatomic gas is supplied heat Q very slowly keeping the pressure constant. The work done by the gas is

A
$$\dfrac{2}{5} Q $$

B
$$\dfrac{3}{5} Q $$

C
$$\dfrac{Q}{5} $$

D
$$\dfrac{2}{3} Q $$

Solution

$$\dfrac {\triangle U }{Q } =\dfrac {3 }{5 } , or\triangle U=\dfrac { 3 }{5 } Q$$
From the first law of thermodynamics
$$Q=\triangle U+W$$
$$W=\dfrac { 2 }{5 } Q$$
Question 10
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$$1\ g$$ of $$H_2O$$ changes from liquid to vapour phase at constant pressure at $$1\ atm$$. The volume increases from $$1\ cc\ to\ 1671\ cc$$. The heat of vaporisation at this pressur is $$540\ cal/g$$. The increase in internal energy of water is

A
$$2099 J$$

B
$$3000 J$$

C
$$992 J$$

D
$$2122 J$$

Solution

Work done is given as-
$$W = P(dV) = 1.01\times 10^5 (1671 - 1)\times 10^{-6} = 167 J$$
$$\Delta Q$$ = $$\Delta U$$ + $$\Delta W$$
or
$$\Delta U$$ = $$\Delta Q$$ - $$\Delta W$$
$$= mL - 167 = (540 \times 4.2) - 167 = 2099 J$$