When 1 kg of ice at 0°C melts to water at 0°C, the resulting change in its entropy, taking latent heat of ice to be 80 cal/°C,is

273 cal/K
8 × 104 cal/K
80 cal/K
293 cal/K

Solution

During an isothermal expansion, a confined ideal gas does –150 J of work against its surroundings. This implies that

(a) 150 J heat has been removed from the gas

(b) 300 J of heat has been added to the gas

(d) 150 J of heat has been added to the gas

Solution

(a, d)

If a process is expansion then work done is positive so answer will be (a).

But in question work done by gas is given –150J so that according to it answer will be (d).

If ΔU and ΔW represent the increase in internal energy and work done by the system respectively in a thermodynamical process, which of the following is true?

ΔU = -ΔW,in an adiabatic process
ΔU = ΔW,in an isothermal process
ΔU = ΔW,in an adiabatic process
ΔU = -ΔW,in an isothermal process

Solution

By first law of thermodynamics,

ΔQ = ΔU + ΔW

In adiabatic process, ΔQ = 0

∴ ΔU = -ΔW

In isothermal process, ΔU = 0

∴ ΔQ = ΔW

One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically.If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be

(T – 4) K
(T + 2.4) K
(T – 2.4) K
(T + 4) K

Solution

For an isothermal expansion of a perfect gas, the value of ^ΔP⁄_P is equal to

-γ^1/2 ^ΔV⁄_V
-^ΔV⁄_V
-γ ^ΔV⁄_V
-γ² ^ΔV⁄_V

Solution

Differentiate PV= constant w.r.t V

⇒ P ΔV + V ΔP = 0

⇒ ^ΔP⁄_P = - ^ΔV⁄_V

A thermodynamic system goes from states (i) P₁, V to 2P₁,V (ii) P, V₁ to P, 2V₁. Then work done in the two cases is

zero, zero
zero, PV₁
PV₁, zero
PV₁, P₁V₁

Solution

Solution

(b,d)

(a)Process is not isothermal.

(b)Volume decreases and temperature decreases

ΔU= negative,

So,ΔQ= negative

(c)Work done in process A → B → C is positve

(d)Cycle is clock wise,so work done by the gas is positive

Figure shows the variation of internal energy (U) with the pressure (P) of 2.0 mole gas in cyclic process abcda. The temperature of gas at c and d are 300 K and 500 K. Calculate the heat absorbed by the gas during the process.

400 R ln 2
200 R ln 2
100 R ln 2
300 R ln 2

Solution

A closed gas cylinder is divided into two parts by a piston held tight. The pressure and volume of gas in two parts respectively are (P, 5V) and (10P, V). If now the piston is left free and the system undergoes isothermal process, then the volumes of the gas in two parts respectively are

2V, 4V
3V, 3V
5V, V
¹⁰⁄₁₁V,²⁰⁄₁₁V

Solution

The volume on both sides will be so adjusted that theoriginal pressure × volume is kept constant as the piston moves slowly (isothermal change)

P5V = P'V'........... (1)

10PV = P'V''........... (2)

From (1) and (2), V'' =2V'

and from V' + V''= 6V

V' = 2V, V''= 4V

In an adiabatic process, the pressure is increased by ²⁄₃%.If γ= ³⁄₂, then the volume decreases by nearl

⁴⁄₉%
²⁄₃%
1%
⁹⁄₄%

Solution

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