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In this page we have *Practice questions with answers on heat and thermodynamics for JEE and Competitive examinations* . Hope you like them and do not forget to like , social shar
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**Question 1:** What increase in radiated power results when the temperature of a blackbody is increased from 7 to 287

^{0} C

a)4

b) 8

c) 16

d) None of these

Solution
$\frac {P(560K)}{(280K)} = (\frac {560}{280})^4 =16$

**Question 2:** The internal energy of an ideal gas depends on

a) Temperature

b) Pressure

c) Volume

d) None of these

Solution
It depend on temperature only

**Question 3:**
All Rod are of equal length and equal area of crossection.Find the temperature at point B

a, 45

b, 51.4

c, 50

d, 51

Solution
. Let T be the temperature at B then total heat flow at B should be zero

[KA(T-0)/L]+ [KA(T - 90)/2L] + [KA(T-90)/2L] = 0

T + (T - 90)/2 + (T - 90)/2 = 0

2T + (T-90) + (T - 90) = 0

2T + T + T = 180

T = 45

**Question 4:**
An ideal gas is taken through a cyclic thermodynamics process through four steps.

The amount of heat involved in the steps are Q

_{1} = 5960 J,Q

_{2} = - 5600 J,Q

_{3}= -3000 J,Q

_{4} = -3600 J

respectively. The corresponding quantities of Internal energy changes are ΔU

_{1} = 3.760 J ,ΔU

_{2} = - 4800 J,ΔU

_{3} = -1800 J,ΔU

_{4} = ?

find the value ΔU

_{4} & net work done

a, 2930 J, 960 J

b, 2830 J, 900 J

C, 2930 J, -960 J

d, -2930 J, 960 J

Solution

Since in cycle process total internal change is zero . ΔU_{1} + ΔU_{2} + ΔU_{3} + ΔU_{4} = 0

3670 - 4800 - 1800 + ΔU_{4} = 0

ΔU_{4} = 2930 J

Now in Cycle

ΔW = ΔQ

= 5960 - 5600 - 3000 + 3600 = 960 J.

Hence answer is a

**Question 5:**
4 moles of an ideal gas undergoes an isothermal expansion at temperature T during which the volume becomes η times.

W -> Workdone by the gas

-> Change in internal energy

a)

b)

c)

d)

Solution
As temperature is constant ,internal energy will remain constant

$W=nRT ln \frac {V_2}{V_1}$

So ans is d

**Question 6** M moles of a ideal polyatomic gas(C

_{v}=7R/2) are in cylinder at temperature T.A heat Q is suuplied to the gas. Some M/3 moles of the gas dissociated into atoms while temperature remains constant. Find the correct relation

a) 3Q=4MRT

b) 2Q=3MRT

c) Q=4MRT

d) 7Q=4MRT

Solution
Internal energy before supplying the heat

$= \frac {7}{2} MRT$

Total no of polyatomic moles after the heat is supplied

=2M/3

Total no of monatomic moles after the heat is supplied

=5M/3

Total internal after the split

$= \frac {7}{2} (\frac {2M}{3})RT + \frac {3}{2} (\frac {5M}{3}) RT$

$= \frac {29}{6} MRT$

So

$Q= \frac {29}{6} MRT - \frac {7}{2} MRT$

3Q=4MRT

Ans is a

**Question 7** Absorbing power of the surface is .7. Transmitting power of the surface is .1

A total heat Q is incident on the surface. Find the heat reflected back?

a)Q/6

b)Q/4

c) Q/2

d) Q/5

Solution
We know that

a+r+t=1

.7+r+.1=1

r=.2

Now

r=Reflected energy/Total energy

So reflected energy=Q/5

**Question 8**
A hole of radius R

_{1} is made centrally in a circular disc of thickness d and radius R

_{2}.The inner surface is maintained at temperature T

_{1} and other surface is maintained at T

_{2} (T

_{1} > T

_{2}).Thermal conductivity of the circular plate is K.

a. Find the temperature as a function of radius from centre

b. Find the heat flow per unit time

Solution
**Assertion and Reason**

a) Statement I is true ,statement II is true ,statement II is correct explanation for statement I

b) Statement I is true ,statement II is true ,statement II is not a correct explanation for statement I

c) Statement I is true, Statement II is false

d) Statement I is False, Statement II is True

**Question 9**
**STATEMENT 1**:The molar specific heat capacity of the ideal gas in isothermal process is infinity

**STATEMENT 2**:Heat transfer is non zero in the Isothermal process

Solution
The molar specfic heat capacity of the ideal gas in isothermal process is infinity as there is no change in temperature

Heat transfer is non zero in the Isothermal process as dU=0 so dQ=dW so dQ not equal to zero

So both the statement is true but statement II is not a correct explanation for statement I

**Question 10**
**STATEMENT 1**:Equation of state for a real gas is (P+a/V

^{2})(V-b)=nRT

**STATEMENT 2**: Molecular attraction is not neglible and the size of molecules are not neglibile in comparison to average separation between them

Solution
Equation of state for a real gas is (P+a/V^{2})(V-b)=nRT.This statement is true

Molecular attraction is not neglible and the size of molecules are not negligible in comparison to average separation between them. This is also true

Equation of state for a ideal gas is PV=nRT

but Equation of state for a real gas is (P+a/V^{2})(V-b)=nRT because Molecular attraction is not neglible and the size of molecules are not neglibile in comparison to average separation between them

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