Thermodynamics Test

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Thermodynamics Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

A gas has pressure P and volume V. It is now compressed adiabatically to $$\dfrac{1}{32}$$ times the original volume. Given that $$(32)^{1.4}= 128$$, the final pressure is ($$\gamma = 1.4$$)

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

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

C
$$32P$$

D
$$128P$$

Solution

$$\displaystyle{PV^{\gamma} = P_1\left(\frac{V}{32}\right)^{\gamma}}$$
$$P_1 = (32)^{\gamma}\\ \Rightarrow P_1 = (32)^{1.4}\\ \Rightarrow P_1 = 128P$$
Question 2
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A polyatomic gas $$\displaystyle{\left(\gamma = \frac{4}{3}\right)}$$ is compressed to $$\displaystyle{\frac{1}{8}^{th}}$$ of its volume adiabatically. If its initial pressure $$P_o$$, its new pressure will be

A
$$8P_o$$

B
$$16P_o$$

C
$$6P_o$$

D
$$2P_o$$

Solution

$$\displaystyle{P_oV^{4/3} = P_1\left(\frac{V}{8}\right)^{4/3} \Rightarrow P_1 = P_o 8^{4/3} = 16 P_o}$$
Question 3
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A sample of gas expands from volume $$V_1$$ to $$V_2$$. The amount of work done by the gas is greatest when the expansion is

A
cannot be infered from the given data

B
isobaric

C
adiabatic

D
equal in all casses

Solution
Question 4
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In pressure-volume diagram, the isochoric, isothermal, isobaric and iso-entropic parts respectively, are

A
BA, AD, DC, CB

B
DC, CB, BA, AD

C
AB, BC, CD, DA

D
CD, DA, AB, BC

Solution

From C to D, V is constant. So process is isochoric.
From D to A, the curve represents constant temperature . So the process is isothermal.
From A to B, Pressure is constant. So, the process is isobaric.
BC represents constant entropy
Question 5
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For an isothermal expansion of a perfect gas, the value of $$\displaystyle{\dfrac{\Delta P}{P}}$$ is equal to

A
$$\displaystyle{\gamma^{1/2} \dfrac{\Delta V}{V}}$$

B
$$\displaystyle{- \dfrac{\Delta V}{V}}$$

C
$$\displaystyle{- \gamma \dfrac{\Delta V}{V}}$$

D
$$\displaystyle{\gamma^{2} \dfrac{\Delta V}{V}}$$

Solution

Differentiate $$PV = constant$$ w.r.t. $$ V$$
$$\Rightarrow$$$$P\Delta V + V\Delta P = 0$$
$$\Rightarrow$$$$\displaystyle{\dfrac{\Delta P}{P} = -\dfrac{\Delta V}{V}}$$
Question 6
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If a gas is heated at constant temperautre its isothermal compressibility

A
remains constant

B
increases linearly with temperature

C
decreases linearly with temperature

D
decreases inversely with temperature

Solution

As we know that isothermal compressibility is given as,
$${K}_{T}=-\dfrac { 1 }{ V } (\dfrac { \delta V }{ \delta P } ) \quad (T=constant)$$
so if it is heated at constant temperature then its isothermal compressibility will also remain constant.
Question 7
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Directions For Questions

A $$2000\ cc$$ sample of $$O_2$$ is confined to a cylinder. Initially, pressure of the gas is $$10^{5} N/m^2$$ and its temperature $$27^{o}C$$. The gas is subjected to a cyclic process. In the first step, its pressure is made twice at constant volume. In the second step, it expands to its initial pressure adiabatically and in the final step, the gas undergoes isobaric compression and finally attains its initial volume. Consider that the gas is ideal.

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Temperature of the at the end of the first step, i.e. just at the start of adiabatic expansion will be -

A
$$177^{o}C$$

B
$$327^{o}C$$

C
$$127^{o}C$$

D
$$148^{o}C$$

Solution

$$P\propto T$$ or $$\displaystyle{\dfrac{P_2}{P_1} = \dfrac{T_2}{T_1}}$$
$$\displaystyle{\dfrac{2P}{P} = \dfrac{T_2}{300 K}}$$
$$T_2 = 600 K$$
Question 8
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Directions For Questions

One mole of an ideal gas is taken from an initial state $$P_o, V_o$$ and temperature $$T_o$$ through the following activities. ($$\gamma = \dfrac{C_p}{C_v}, C_p = \dfrac{7R}{2}, C_v = \dfrac{5R}{2}$$)
(i) Heating at constant volume to a temperature three times.
(ii) Adiabatic expansion to a volume $$2V_o$$
(iii) Cooling at constant volume to a temperature on third
(iv) Adiabatic compression so that it is returned to its initial state

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The cycle on a $$P V$$ diagram is represented as

A

B

C

D
none of these

Solution

plot the given condition on $${P}-{V}$$ graph ,we will find the answer
now by keeping the volume constant we are heating the process $${A-B}$$ and temperature will reach from $${T}_{0}\ to\ {3T}_{0}$$
as we know $${PV}={nRT}$$ hence for isochoric process $$P \propto {T}$$
as pressure increases temperature will also increases.
for $${B}-{C}$$ adiabatic expansion from ($${V}_{0}-{2V}_{0}$$) and heat $${Q}=0$$
again from point c to d isochoric cooling will occur hence pressure will reduce along with temperature and reach at $${P}_{0}$$ and volume $${2{V}_{0}}$$
form D-A it will achieve it's original state by compressing the gas at constant $${Q}=0$$
Question 9
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In an adiabatic process, the pressure is increased by $$\displaystyle{\dfrac{2}{3}}$$%. If $$\displaystyle{\gamma = \dfrac{3}{2}}$$, then the volume decreases by nearly

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

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

C
$$1\%$$

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

Solution

$$PV^{3/2}$$ = K
$$\log P + \displaystyle{\dfrac{3}{2}}\log V = \log K$$

$$\displaystyle{\dfrac{\Delta P}{P} + \frac{3}{2}\dfrac{\Delta V}{V}}$$ = 0
$$\displaystyle{\dfrac{\Delta V}{V} = -\dfrac{2}{3}\dfrac{\Delta P}{P}}$$
or $$\dfrac{\Delta V}{V} = -\dfrac{2}{3} \times \dfrac{2}{3}$$
$$=\displaystyle{-\dfrac{4}{9}}$$
Question 10
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Two moles of helium gas are taken over the cycle ABCDA, as show in the P - T diagram. Assuming the gas to be ideal the work done on the gas in taking it from state A to B is :

A
$$200 R$$

B
$$300 R$$

C
$$400 R$$

D
$$500 R$$

Solution

Work done in $$A\rightarrow B$$ Isobaric process : $$W = P\Delta V$$

$$W = \mu R \Delta T$$

$$W = 2 \times R \times [T_
2 - T_1]$$

$$W = 2 \times R \times [500 - 300]$$

$$\Rightarrow$$$$W = 400R$$