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First Law of ...

First Law of Thermodynamics

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Q.

Assuming that water vapour is an ideal gas, the internal energy change (∆U) when 1 mol of water is vapourised at 1 bar pressure and 100 $^{\circ}$ C, (given : molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol $^{- 1}$ and R = 8.3 J mol $^{- 1}$ K $^{- 1}$ ) will be

41.00 kJ mol $^{- 1}$
4.100 kJ mol $^{- 1}$
4.100 kJ mol $^{- 1}$
4.100 kJ mol $^{- 1}$

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Q.

What violates the second law of thermodynamics?

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Q. In a cyclic process, heat transfers are

+ 14.7 k J

,

- 25.2 k J

,

- 3.56 k J

and

+ 31.5 k J

. What is the net work done in the net cyclic process?

$+ 10.56 k J$
$- 10.56 k J$
$- 17.44 k J$
$+ 17.44 k J$

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Q.

A vessel contains 100 litres of a liquid $x$ . Heat is supplied to the liquid in such a fashion that, Heat given = change in enthalpy. The volume of the liquid increases by 2 litres. If the external pressure is one atm, and 202.6 Joules of heat were supplied then, [ U - total internal energy ]

∆U = 0, ∆H = 0
∆U = + 202.6 J, ∆H = + 202.6 J
∆U = - 202.6 J, ∆H = - 202.6 J
∆U = 0, ∆H = + 202.6 J

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Q.

The entropy change in an adiabatic process is zero.

True
False

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Q. A carnot engine works between 400k and 600k if it absorbs 3 × 10 ^6 cal heat at highest temperature how much work per cycle can engine perform

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Q. A gas at a pressure of 1 atm is compressed at a cons†an t pressure from 10 litre to 2 litre. If 500J of heat leaves the gas, claculate the work done on the gas and the change in the internal energy

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Q.

Is heat a parabolic equation?

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Q. The limitation of the first law of themodynamics is:

It does not indicate the possibility of a spontaneous process proceeding in a definite direction
It assigns a quality to different forms of energy
Indicates the direction of any spontaneous process
None of the above

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Q. WHY IS WORK OBTAINED MAXIMUM IN REVERSIBLE AND MINIMUM IN IRREVERSIBLE PROCESS

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Q. One mol of an ideal monoatomic gas at temperature T and volume 5 L expands to 10 L against a constant external pressure of 2 atm under adiabatic conditions. Then the final temperature of the gas will be:

$T + \frac{30}{2 R}$
$T - \frac{20}{3 R}$
$T + \frac{20}{3 R}$
$\frac{T}{2^{\frac{5}{3}} + 1}$

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Q. A container with insulating walls is divided into two equal parts by a partition fitted with a valve. One part is filled with an ideal gas at a pressure P and temperature T, whereas the other part is completely evacuated. If the valve is suddenly opened, the pressure and temperature of the gas respectively will be:

$\frac{P}{2}, T$
$\frac{P}{2}, \frac{T}{2}$
$P, T$
$P, \frac{T}{2}$

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Q. The entropy change in an adiabatic process is:

zero
positive
negative
remains same

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Q. The work done by a closed system in a reversible process in case of expansion is always ___ that done in an irreversible process.

less than or more than
equal to
less than
more than

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Q. A gas is cooled and loses

65 J

of heat. The gas contracts as it cools and the work done on the system is equal to

20 J

. What is

△ U

?

$- 45 J$
$- 65 J$
$+ 45 J$
$- 20 J$

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Q. A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy

Δ U

of the gas in joules will be

– 500J
– 505J
+ 505J
1136.25J

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Q. Assuming that, water vapour is an ideal gas, the internal energy change

(Δ U

) when

1

mol of water is vaporised at

1

bar pressure and

100^{o} C

, (given: molar enthalpy of vaporisation of water at

1

bar and

373 K = 41 k J {m o l}^{- 1}

and

R = 8.314 J K^{- 1} {m o l}^{- 1}

) will be:

$41.00 J {m o l}^{- 1}$
$4.100 J {m o l}^{- 1}$
$3.7904 J {m o l}^{- 1}$
$37.904 J {m o l}^{- 1}$

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Q.

For the process $H_{2} O (l) (1 b a r, 373 K) \to H_{2} O (g) (1 b a r, 373 K)$ , the correct set of thermodynamic parameters is

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Q. Two mole of ideal diatomic gas

(C_{v, m} = \frac{5}{2} R)

at 300 K and 5 atm expanded irreversly and adiabatically to a final pressure of 2 atm against a constant pressure of 1 atm. Calculate q, w,

Δ H

and

Δ U .

$w = - 1246.1 J, q < 0, Δ H = - 1745.9 J, Δ U = 0$
$w = 1745 J, q = 0, Δ H = 1247 J, Δ U = 0$
$w = - 1246 J, q = 0, Δ H = - 1746 J, Δ U = - 1247 J$
$w = - 1745.6 J, q > 0, Δ H = - 1245 J, Δ U = - 1245 J$

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Q. In an adiabatic process which of the following is true?

$q = w$
$q = 0$
$Δ E = q$
$P Δ V = 0$

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Q. what is the sign of work done in reversible and irreversible reaction?? why we take work done by the systen as positive in the equation delH= delQ delW and work done by system as negative when we use -p(delv) in which finally we get work fone by system as negative

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Q. For the reaction,

2 C O + O_{2} \to 2 C O_{2}; Δ H = - 560 K J

.

Two moles of CO and one mole of

O_{2}

are taken in a container of volume 1 L. They completely form two moles of

C O_{2}

the gases deviate appreciably from ideal behaviour. If the pressure in the vessel changes from 70 to 40 atm, then the magnitude (absolute value) of

Δ U

at 500 K is

x k J

.

The value of

0.01 x

is ______(nearest integer).

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Q. A thermodynamic system goes from states (i)

P_{1}, V

to

2 P_{1}, V

and

(i i) P, V_{1} t o P, 2 V_{1}

The work done in the two cases respectively is:

Zero, Zero
Zero, $- P V_{1}$
$- P V_{1},$ Zero
$- P V_{1}, - P_{1} V_{1}$

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Q.

If the first law of thermodynamics is represented by $Δ u = q + w$ . The $q$ & $w$ represent?

Work done on the system & heat added to the system respectively
Work done by the system & heat added to the system respectively
Heat added to the system & work done on the system respectively
Heat take from the system & work done on the system respectively

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Q. In an isobaric reversible process, the ratio of heat supplied to the system (

d q

) and work done by the system (

d W

) for a ideal monoatomic gas is:

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Q. If the amount of heat given to a system be

85 J

and the amount of work done by the system be

35 J

, then the change in the internal energy of the system is

$50$
$120$
$55$
$160$

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Q. An ideal gas in thermally insulated vessel at internal pressure

= p_{1}, v o l u m e = V_{1}

and absolute temperature

= T_{1}

expands irreversibly against zero external pressure, as shown in the diagram.
The final internal pressure, volume and absolute temperature of the gas are

p_{2}, V_{2} a n d T_{2},

respectively. For this expansion

q = 0
$T_{2} = T_{1}$
$p_{2} V_{2} = p_{1} V_{1}$
$p_{2} V_{2}^{γ} = p_{1} V_{1}^{γ}$

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Q. Match the column

\begin{matrix} Column - I & Column - II (a) & Reversible cooling of an ideal gas at & (P) & w = 0; q < 0; Δ U < 0 constant volume (b) & Reversible isothermal expansion of an & (Q) & w < 0; q > 0; Δ U > 0 ideal gas (c) & Adiabatic expansion of non-ideal gas into & (R) & w = 0; q = 0; Δ U = 0 vacuum (d) & Reversible melting of sulphur at normal & (S) & w < 0; q > 0; Δ U = 0 melting point \end{matrix}

a - P; b - S; c - R; d - Q
a - P; b - Q; c - S; d - R
a - R; b - S; c - Q; d - P
a - Q; b - R; c - S; d - P

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Q. A gas in the initial state A having pressure

{5 N m}^{- 2}

and volume

{2 m}^{3}

is changed to state C having pressure

{15 N m}^{- 2}

and volume

{6 m}^{3}

following paths ABC and AC. Path ABC includes path AB and BC. The state B has pressure

{15 N m}^{- 2}

and volume

{2 m}^{3}

. Calculate:
(a) Path along which work done is least.
(b) Internal energy at C, if the internal energy of gas at A is 10J and amount of heat supplied to change its state to C through the path AC is 200J.
(c) Amount of heat supplied to gas to go from A to B, if internal energy of gas at state B is 10J.
Find the value of

W_{A C} + U_{C} + Q_{A B}

.

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Q. when a tyre burst, on touching it why feel cold sensation?

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