## Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

## Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

## Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

# Thermodynamics

#### Multiple Choice Questions

1.

The heat of neutralization of a strong base and a strong acid is 57 kJ. The heat released when 0.5 mole of HNO3 solution is added to 0.20 moles of NaOH solution, is :

• 11.4 kJ

• 34.7 kJ

• 23.5 kJ

• 58.8 kJ

A.

11.4 kJ

Heat of neutralization ($∆$H)= 57 kJ, mole of HNO3 = 0.5 mole and mole of NaOH = 0.2 mole. When HNO3 solution is added to NaOH solution, then 0.2 mole of HNO3 solution will combine with 0.2 mole of OH- ions of NaOH solution.

$\therefore$ Heat released= $∆$H x 0.2 = 57 x 0.2 = 11.4 kJ.

2.

The latent heat of vapourisation of water is  2240. If the work done in the process of vapourisation of 1 g is 168 J, then increase in internal energy is

• 2408 J

• 2072 J

• 2240 J

• 1904 J

B.

2072 J

Latent heat of vapourisation of water ( L ) = 2240 J

Mass of water (m)  = 1 g

Work done (dW ) = 168 J

From first law of thermodynamics, heat supplied in vapourisation

(dQ) = mL

= dU + dW

1 × 2240 = dU + 168

( where dU = increase in internal energy )

3.

Helium at 270C has a volume 8 litres. it is suddenly compressed to a volume of 1 litre. The temperature of the gas will be

• 93270C

• 12000C

• 9270C

• 1080C

C.

9270C

${{\mathrm{TV}}_{1}}^{\mathrm{\gamma }-1}={{\mathrm{TV}}_{2}}^{\mathrm{\gamma }-2}\phantom{\rule{0ex}{0ex}}{\mathrm{T}}_{2}=\left(\frac{{\mathrm{V}}_{1}}{{\mathrm{V}}_{2}}\right){\mathrm{T}}_{1}\phantom{\rule{0ex}{0ex}}{\mathrm{T}}_{2}={\left(\frac{{\mathrm{V}}_{1}}{{\mathrm{V}}_{2}}\right)}^{\frac{5}{3}-2}×300\phantom{\rule{0ex}{0ex}}{\mathrm{T}}_{2}={\left(\frac{8}{1}\right)}^{\frac{2}{3}}×300=4×300=1200\mathrm{k}\phantom{\rule{0ex}{0ex}}{\mathrm{T}}_{2}=1200-273={927}^{0}\mathrm{C}$

4.

An engine is working. It takes 100 calories of heat from source and leaves 80 calories of heat to sink. If the temperature of source is 127°C, then temperature of sink is

• 147°C

• 47°C

• 100°C

• 47 K

B.

47°C

Heat taken from source  Q1 = 100 cal

Heat left to sink   Q2 = 80 cal

∴  efficiency of the engine

η = 1

= 1

η = 20 %

Temperature of the source

T1 = 127°C =  400 K

Temperature of the sink

T2 = ?

We know that

η = 1

0.2  = 1

⇒   $\frac{{\mathrm{T}}_{2}}{400}$ = 1 $-$ 0.2

= 0.8

T2 = 0.8 × 400

T2 = 320 K

⇒   T2 = 47°C

5.

Assertion: Mass and volume are extensive properties.
Reason: Mass/volume is also an extensive parameter.

• If both assertion and reason are true and reason is a correct explanation of the assertion.

• If both the assertion and reason are true but the reason is not a correct explanation of the assertion.

• If the assertion is true but the reason is false.

• If both the assertion and reason are false.

C.

If the assertion is true but the reason is false.

Extensive properties are dependent upon the amount of the substance. e.g. mass, volume, etc.

Intensive properties are independent of the amount of the substance e.g. temperature, density.

6.

Anatomy and Bismuth are usually used in thermocouple, because

• a constant e..m.f is produced

• higher thermo e.m.f is produced

• a negative e.m.f is produced

• lower thermo e.m.f is produced

B.

higher thermo e.m.f is produced

When two wires of dissimilar metals like Antimony and Bismuth are joined to form a closed circuit with two junctions and  if the junctions are at different temperatures emf is generated and current flows through the circuit.

7.

Air is expanded from 50 litres to 150 litres at atmosphere the external work done is (1 atmospheric pressure =1$×$${10}^{5}\mathrm{N}/{\mathrm{m}}^{2}$)

• $2×{10}^{-8}\mathrm{J}$

• $2×{10}^{4}\mathrm{J}$

• 200J

• 2000J

B.

$2×{10}^{4}\mathrm{J}$

8.

The heat produced in a long wire is characterized by resistance, current and time through which the current passes. If the errors in measuring these quantities are respectively 1 %, 2% and 1%, then the total error in calculating the energy produced is

• 4%

• 6%

• 4/3 %

• 8%

B.

6%

The heat produced in a wire due to current flown is given by

H = I2Rt

∴

= 2 ×  0.02 + 0.01 + 0.01

$\frac{∆\mathrm{H}}{\mathrm{H}}$ = 6%

# 9.The density of a substance  at  0°C  is 10 g/cc and at  100° C, its density is  9.7  g/cc. The coefficient of linear expansion of the substance is 10-4 102 10-3 102

A.

10-4

Initial temperature ( T) = 0o C

Initial density ( ρ1 ) = 10 g/cc

Final temperature ( T2 ) = 100oC

Final density ( ρ2 )  = 9.7 g/cc

Coefficient of volumetric expansion

( ${\mathrm{\alpha }}_{\mathrm{v}}$ )  =

=

=

= $\frac{0.3}{1000}$

$\mathrm{\alpha }$v = 3 × 10-4

Therefore linear coefficient of linear expansion

$\mathrm{\alpha }$1$\frac{\mathrm{\gamma }}{3}$

=

$\mathrm{\alpha }$1 = 10-4

10.

Assertion: During an adiabatic process, heat energy is not exchanged between the system and its surroundings.

Reason: The temperature of a gas increases when it undergoes an adiabatic expansion.

• If both assertion and reason are true and the reason is a correct explanation of the assertion.

• If both the assertion and reason are true but the reason is not a correct explanation of the assertion.

• If the assertion is true but the reason is false.

• If both the assertion and reason are false.

C.

If the assertion is true but the reason is false.

In the adiabatic process, no heat enters or leaves the system. The system is completely insulated from its surroundings. From the first law of thermodynamics, change in internal energy,

$∆$E = q-W = 0-W = -W

Since the work is done at the expense of internal energy, therefore internal energy of the system decreases and hence temperature of the gas falls.