Thermodynamics Test

Result

Thermodynamics Test - 63

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Question 1
1 / -0

In which condition the relation $$\Delta H=\Delta U+P\Delta V$$ can be said to be true for a closed system?

A
Constant temperature

B
Constant pressure

C
Both Constant temperature and pressure together

D
Constant temperature and volume

Solution
Question 2
1 / -0

For an adiabatic expansion of a perfect gas, the value of $$\Delta P/P$$ is equal to:

A
$$\cfrac{\Delta V}{V}$$

B
$$\gamma \cfrac{\Delta}{V}$$

C
$$-\gamma \cfrac{\Delta}{V}$$

D
$$-{\gamma}^{2} \cfrac{\Delta}{V}$$

Solution

$$\begin{array}{l} For\, \, an\, \, adiabatic\, \, change\, to\, \, a\, perfect\, \, gas\, : \\ P{ V^{ \gamma } }=K\, \, \, \, .................\left( 1 \right) \\ =K{ V^{ -\gamma } } \\ Differentiating\, \, \, with\, \, respect\, \, to\, \, V: \\ \frac { { dP } }{ { dV } } =-\gamma K{ V^{ \left( { -\gamma -1 } \right) } }=\frac { { { V^{ -\gamma } } } }{ V } \, \, \, \, .............\left( 2 \right) \\ { V^{ \left( { -\gamma -1 } \right) } }=\frac { { { V^{ -\gamma } } } }{ V } \\ so\, \, equation\, \, \left( 2 \right) \, \, can\, \, be\, written: \\ \frac { { dP } }{ { dV } } =-\gamma \frac { { K{ V^{ -\gamma } } } }{ V } ............\left( 3 \right) \\ form\, \, equation\, \left( 1 \right) \\ { V^{ \gamma } }=\frac { K }{ P } \\ { V^{ -\gamma } }=\frac { P }{ K } \\ This\, \, can\, \, be\, \, ressed\, \, as \\ \frac { { dP } }{ P } -\gamma \frac { { dV } }{ V } \\ providing\, \, \Delta p\, \, and\, \, \Delta V\, \, are\, \, sufficiently\, \, small,it\, follows; \\ \frac { { \Delta P } }{ P } =-\gamma \frac { { \Delta V } }{ V } \end{array}$$
Question 3
1 / -0

An ideal gas at a pressure of 1 atmosphere and temperature of $$27 ^\circ C$$ is compressed adiabatically until its pressure becomes 8 time the initial pressure, then the final temperature

A
$$627 ^\circ C$$

B
$$527 ^\circ C$$

C
$$427 ^\circ C$$

D
$$327 ^\circ C$$

Solution
Question 4
1 / -0

The pressure and densitey of adiatomicgas (y=7/5) change adiabatically form (p,D) to (p,D) if

A
1/128

B
32

C
128

D
None of above
Question 5
1 / -0

During an adiabatic process,the cube of the pressureis found to be inversely proportional to the fourth power of the volume. Then, the ratio of specific heats is :-

A
1

B
1.33

C
1.67

D
1.4

Solution

$$\begin{array}{l} Equation\, of\, an\, adiabatic\, process\, is\, p{ V^{ \gamma } }=cons\tan t\, \, \, ---\left( 1 \right) \\ Given, \\ { p^{ 3 } }=\frac { k }{ { { V^{ 4 } } } } { p^{ 3 } }{ V^{ 4 } }=k\, \, \, \, \, \, ---\left( 2 \right) \\ Comparing\, equation\, \left( 1 \right) \, and\, \left( 2 \right) \, we\, get \\ \gamma =\frac { 4 }{ 3 } =1.33 \\ Hence,\, the\, option\, B\, is\, the\, correct\, answer. \end{array}$$
Question 6
1 / -0

Consider one-gram molecule of a gas initially at $$147 ^ { \circ } C$$ expands isothermally till the point where the volume becomes $$10$$ times. How much is the amount of heat absorbed ?

A
$$920{ cal }$$

B
$$585{ cal }$$

C
$$1911{ cal }$$

D
$$1542{ cal }$$

Solution
Question 7
1 / -0

It is required to double the pressure of a gas contained in a steel cylinder by heating. If the initial temperature $$27^\circ C$$ upto which it the gas is to be heated is

A
$$327^{ \circ }C\\ $$

B
$$273^{ \circ }C$$

C
$$108^{ \circ }C$$

D
$$58^{ \circ }C$$

Solution
Question 8
1 / -0

Which of the following is not thermodynamical function

A
Enthalpy

B
Work done

C
Gibb's energy

D
Internal energy

Solution
Question 9
1 / -0

A gas expands from 40 litres to 90 litres at a constant pressure of 8 atmosphere work done by the gas during expansion

A
$$ 4 \times 10^{-4} J $$

B
$$ 4 \times 10^{4} J $$

C
$$ 4 \times 10^{3} J $$

D
$$ 4 \times 10^{2} J $$

Solution
Question 10
1 / -0

An ideal gas is taken through a cyclic thermo dynamical process through four steps. The amounts of heat involved in steps are
$$Q_1 = 5960 J, Q_2 = -5585 J, Q_3 = -2980 J, Q_4 = 3645 J$$; respectively, the corresponding works involved are $$W_1 = 2200 , W_2 = -825 J, W_3 = -1100 J$$ and $$W_4$$ respectively. Find the value of $$W_4$$ and efficiency of the cycle

A
1315 J, 10%

B
765 J, 11%

C
both of them

D
none of these

Solution

Given;-
$$\begin{matrix} { Q_{ 1 } }=5960J\, \, \, ,{ W_{ 1 } }=2200J \\ { Q_{ 2 } }=5585J\, \, \, ,{ W_{ 2 } }=825J \\ { Q_{ 3 } }=2980J\, \, \, ,{ W_{ 3 } }=1100J \\ { Q_{ 4 } }=3645J\, \, \, ,{ W_{ 4 } }=find \\ \eta =find \\ \end{matrix}$$
In cycic process $$\Delta U=0$$
$$A/q$$ $$1st$$ law of thermodynamics
$$\Delta Q= \Delta U+W$$
$$\Rightarrow +1040 = 275+W_4$$
$$\Rightarrow W_4=1040-275$$
$$=765$$
$$\eta \% = \dfrac{{output}}{{input}} \times 100$$

$$ = \dfrac{{1040}}{{5960 + 3645}}$$

$$ = \dfrac{{208}}{{1321}} \times 100$$
Hence, we get
$$\eta = 10.82\% $$