**CHAPTER 03 **

**THE FIRST LAW OF THERMODYNAMICS **

**THE FIRST LAW OF THERMODYNAMICS**

**3-1 THE FIRST LAW OF THERMODYNAMICS**

**Problem 3-1**

**A system receives **200 *calories***of heat and work done by the **
**system is**736 *joules***. What is change in its internal energy? **

) 186 . 4 1

( *calorie*= *J* **. B.U. B.Sc. 2000A **

**Solution **

According to first law of thermodynamics
*dQ*=*dU*+*dW*

*dU* =*dQ*−*dW*

*J*

*dU* =(200)(4.186)−736=101.2

**Problem 3-2 **

**A system received **1254 *joules***of heat. Calculate the work **
**done by the system if the change in its internal energy is **

*calories*

200 **. B.U. B.Sc. 2007A **
**Solution **

According to first law of thermodynamics
*dQ*=*dU*+*dW*
*dW* =*dQ*−*dU*
Now *dQ*=1254 *J*
*J*
*J*
*calories*
*dU* =200 =(200)(4.186) =837.2
Hence
*dW* =1254−837.2=416.8 *J*

**Problem 3-3 **

**A system receives **150**calories of heat and the work done by **
**the system is **418**joules. What is the change in its internal **
**energy (**1 *calorie*=4.18 *J***) B.U. B.Sc. 2009S **
**Solution **

According to the first law of thermodynamics
*dW*
*dU*
*dQ*= +
*dW*
*dQ*
*dU* = −
Now *dQ*=150 *calories*=(150)(4.18)=627 *J*
*dU* =418 *J*
Therefore
*J*
*dU* =627−418=209
**Problem 3-4 **

**An ideal gas expands isothermally, performing **5.00×103*J*
**of work in the process. Calculate **

**(a) the change in internal energy of the gas and **
**(b) the heat absorbed during this expansion. **
**Solution **

(a) *dU* =*dEINTERNAL* =0 because there is no change in
temperature.

(b) *dQ*=*dU* +*dW*

*J*
*dQ*=0+(5.00×103)=5.00×103
**Problem 3-5 **

**Let **1.00 *kg*** of water be converted to steam by boiling. The **
**volume changes from an initial value of **1.00×10−3*m*3**as a **
**liquid to **1.671 *m*3**as steam. For this process, find **

**(a) the work done on the system **
**(b) the heat added to the system and **

**(c) the change in internal energy of the system. **

**(Heat of vapourization ** =2.26×106*J* *kg*−1**, 1 atmosphere **

2
3
10
013
.
1 × *N* *m*− **) K.U. B.Sc. 2003 **
**Solution **

*dW* = *p* *dV*
*J*
*dW* =(1.013×105){1.671−(1.00×10−3)}=1.692×105
(b) *dQ*=*m* *L _{v}* =(1)(2.26×106)=2.26×106

*J*(c)

*dU*=

*dQ*−

*dW*

*dU*=(2.26×106)−(1.692×105)=2.091×106

*J*

**Problem 3-6**

**A ** 40 *W* **heat source is applied to a gas sample for**25 *s***, **
**during which time the gas expands and does **750 *J* **of work **
**on its surroundings. By how much does the internal energy **
**of the gas change? **

**Solution **

According to the first law of thermodynamics
*dQ*=*dU*+*dW*
*dU* =*dQ*−*dW*
Now
*dQ*=*P*∆*t* =(40)(25)=1000 *J*
*dW* =750 *J*
Hence
*dU* =1000−750=250 *J*

**3-2 HEAT CAPACITIES OF AN IDEAL GAS **
**Example 3-7 **

**In an experiment, **1.35 *mol***of oxygen (****O ) are heated at **_{2}

**constant pressure starting at**110*C***. How much heat must be **
**added to the gas to double its volume? Given that **

1
1
4
.
29 − −
= *J* *mol* *K*
*CV* **is for oxygen. **
**Solution **

According to Charles’s law
= =
2
2
1
1
*T*
*V*
*T*
*V*
constant
1
1
1
1
1
1
2
2 2
2
*T*
*T*
*V*
*V*
*T*
*V*
*V*
*T* _{} =
=
=

The amount of heat added to the system at constant pressure is
given by
*Q*=*n* *C _{P}*

*dT*=

*n*

*C*(

_{P}*T*

_{2}−

*T*

_{1})=

*n*

*C*(2

_{P}*T*

_{1}−

*T*

_{1})

*J*

*T*

*C*

*n*

*Q*

*4 1 =(1.35)(29.4)(11+273)=1.127×10 =*

_{P}**Example 3-8**

**Twelve grams of nitrogen **(*N*_{2})**in a steel tank are heated **
**from **250*C*** to **1250*C***. (a) How many moles of nitrogen are **
**present? (b) How much heat is transferred to the nitrogen? **
**(The specific heat of ** **N at constant volume is **_{2}

1
1
8
.
20 *J* *mol*−*kg*− **). **
**Solution **

(a) 28 grams of *N*_{2} is = 1 mol

12 grams of **N **_{2} is = =0.429 *mol* =*n*
28
12
(b) *Q*=*n* *C _{V}*

*dT*

*J*

*Q*=(0.429)(20.8){(125+273)−(25+273)}=892

**Example 3-9 **

**The mass of helium atom is ** 6.66×10−27*kg***. Compute the **
**specific heat at constant volume for helium gas **
**(in***J* *kg*−1*K*−1**) from the molar heat capacity at constant **
**volume. (Given that ***CV* =12.5 *J* *mol*−1*K*−1**). **

**Solution **

Now *C _{V}* =12.5

*J*

*mol*−1

*K*−1

The mass of one mole of helium gas is
(mass of helium atom) *N _{A}*

*kg*
3
23
27
10
011
.
4
)
10
022
.
6
)(
10
66
.
6
( × − × = × −
=
Hence
1
1
3
3 3.116 10
10
011
.
4
5
.
12 − −
− = ×
×
= *J* *kg* *K*
*CV*
**Example 3-10 **

**Propane gas **(*C*_{3}*H*_{8})**behaves like an ideal gas with**γ =1.127**. **
**Determine the molar heat capacity at constant volume and **
**the molar heat capacity at constant volume. **

**Solution **
We know that
*R*
*C*
*CP* = *V* +
But
*V*
*P*
*C*
*C*
=

γ or *CP* =γ *CV* , therefore above expression

becomes
γ *C _{V}* =

*C*+

_{V}*R*(γ − )1

*CV*=

*R*1 1 5 . 65 1 127 . 1 314 . 8 1 − − = − = − =

*R*

*J*

*mol*

*K*

*C*γ

_{V}*C*=

_{P}*C*+

_{V}*R*=65.5+8.314=73.8

*J*

*mol*−1

*K*−1

**Example 3-11 **

**The heat capacity at constant volume of a certain **
**amount of a monoatomic gas is **49.8 *J*/*K***. **
**(a) Find the number of moles of the gas. **

**(b) What is the internal energy of the gas at ***T* =300 *K***? **
**(c) What is the heat capacity of the gas at constant **

**pressure? **
**Solution **

(a) For monoatomic gas
*CV* *n* *R*
2
3
=
*moles*
*R*
*C*
*n* *V*
4
)
314
.
8
(
5
)
8
.
49
(
2
3
2
≅
=
=
(b) *U* *CVT* *J*
4
10
494
.
1
)
300
)(
8
.
49
( = ×
=
=
(c) *C _{P}*

*C*

_{V}*n*

*R*

*C*

_{V}*C*3 2 + = + =

_{V}*C*

_{V}*n*

*R*2 3 = Θ

*CP*

*CV*(49.8) 83

*J*/

*K*3 5 3 5 = = =

**Example 3-12**

**One mole of a monoatomic ideal gas is initially at **273 *K*
**and one atmosphere. **

**(a) What is its initial internal energy? **

**(b) Find its final internal energy and the work done by the **
**gas when **500 *J***of heat are added at constant pressure. **
**(c) Find the same quantities when **500 *J***of heat are added **
**at constant volume. **

**Solution **

(a) The initial internal energy is given by
*U _{i}*

*C*

_{V}T*R*

*T*2 3 = =

*J*

*U*(8.314)(273) 3405 2 3 = =

_{i}(b) Now *dQ*=*C _{P}dT* or

*P*

*C*

*dQ*

*dT*=

*J*

*C*

*C*

*dQ*

*C*

*dQ*

*C*

*dT*

*C*

*dU*

*V*

*P*

*P*

*V*

*V*300 ) 3 / 5 ( 500 ) / ( = = = = =

Therefore the final internal energy and work are given by
*U _{f}* =

*U*+

_{i}*dU*=3405+300=3705

*J*

*dW* =*dQ*−*dU* =500−300=200 *J*

**3-2 WORK DONE ON OR BY AN IDEAL GAS **
**3-2(A) WORK DONE AT CONSTANT VOLUME **
**Example 3-13 **

**Calculate the increase in internal energy of ten grams of **
**oxygen whose temperature is increased by **100*C*** at constant **
**volume. Given that ***C _{v}* =1.82

*J*

*g*−1 0

*C*−1

**.**

**Solution **

According to first law of thermodynamics
*dQ*=*dU*+*dW*
** ***m* *C _{v}dT* =

*dU*+

*p*

*dV*) 0 ( ) 10 )( 82 . 1 )( 10 (

*g*

*J*

*g*−1 0

*C*−1 0

*C*=

*dU*+

*p*182

*J*=

*dU*

**3-2(B) WORK DONE AT CONSTANT PRESSURE **
**Problem 3-14 **

**A gas expands at atmospheric pressure and its volume **
**increases by ***500 cm*3**. Calculate the work done by the gas. **
**(**1** Atmosphere** 5 2
/
10
013
.
1 × *N* *m*
= **) B.U. B.Sc. 1998A **
**Solution **

The work done by the gas at constant pressure is given by
*W* = *p* *dV*

*J*
*W* _{=}(1.013_{×}105)(500_{×}10−6)_{=}50.65
**Problem 3-15 **

**One kilogram water is converted to steam at standard **
**atmospheric pressure. The volume changes from **1×10−3*m*3
**as a liquid to **1.671 *m*3**as steam. For this process calculate **
**the work done on the system when the pressure is **

2
5
/
10
013
.
1 × *N* *m* ** . K.U. B.Sc. 1998 **
**Solution **

The work done by the vapourizing water is
*W* = *p* *dV* = *p* (*V _{f}* −

*V*)

_{i}*J*

*W*5 3 5 10 691 . 1 ) 10 1 671 . 1 )( 10 013 . 1 ( × − × = × = −

**Problem 3-16**

**A gas expands at atmospheric pressure and its volume **
**increases by***334 cm*3**. Find the work done by the gas. The **
**atmospheric pressure is **1.013×106*dynes*/*cm*2**. **

** B.U. B.Sc. (Hons.) 1988A **
The work done by the gas, at constant pressure, is given by

)
(*Vf* *Vi*
*p*
*dV*
*p*
*W* = = −
Now
** ** 6 2
/
10
013
.
1 *dynes* *cm*
*p*= ×
_{2} _{2} 2
5
6
/
)
10
(
10
)
10
013
.
1
(
*m*
*N*
*p* _{−}
−
×
×
= *dyne* 5*N*
10
1 = −
Θ

*p*=1.013×105*N*/*m*2
3
4
3
3
2
3
10
34
.
3
)
10
)((
334
(
334 *cm* *m* *m*
*dV* = = − = × −
Hence
*J*
*dW* =(1.013×105)(3.34×10−4)=33.834
**Problem 3-17 **

**A gas is compressed at a constant pressure of **0.8 *atm*** from **
*litres*

9 **to**2 *litres***. In the process, **400 *J***of energy leaves the **
**gas by heat. **

**(a) What is the work done on the gas? **

**(b) What is the change in its internal energy? **
**Solution **

(a) The work done is given by
*dW* = *p* *dV*
*dW* {(0.8) (1.013 105)}{(2 9) 10 3} 567.28 *J*
−
=
×
−
×
×
= −

(b) The desired change in internal energy of the system is given by

*dQ*=*dW* +*dU*
*dU* =*dQ*−*dW*

*J*

**3-2(C) WORK DONE AT CONSTANT TEMPERATURE **
**Problem 3-18 **

**A gram molecule of a gas at **770*C*** expands isothermally to **

**its double volume. Calculate the amount of work done. **
** B.U. B.Sc. (Hons.) 1987A **

**Solution **

The work done is given by

_{}
=
*i*
*f*
*V*
*V*
*n*
*T*
*R*
*n*
*W* λ
*J*
*V*
*V*
*n*
*W* (1)(8.314)(77 273) 2 2017
0
0 _{=}
+
= λ
**Problem 3-19 **

**Calculate the work done by an external agent in **
**compressing ** 1.12 *moles*** of oxygen from volume of **

*litres*

4 .

22 **and **1.32 *atm*** pressure to **15.3 *litres***at the same **
**temperature. P.U. B.Sc. 2001 **
**Solution **

The desired work on the gas by the external agent is given by

_{}
=
=
*i*
*f*
*i*
*i*
*i*
*f*
*V*
*V*
*n*
*V*
*p*
*V*
*V*
*n*
*T*
*R*
*n*
*W* λ λ
×
×
×
= −
4
.
22
3
.
15
)
10
4
.
22
)}(
10
013
.
1
(
32
.
1
( 5 3 *n*
*W* λ
*W* =−1.142×103*J* =−1.142 *kJ*
**Problem 3-20 **

**A sample of gas consisting of ** 0.11 *moles*** is compressed **
**from a volume of ** 4.0 *m*3** to ** 1.0 *m*3** while its pressure **
**increases from **10** to ** 40 *Pa* (*N*/*m*2)**. Calculate the work **
**done. P.U. B.Sc. 2002, K.U. B.Sc. 2008 **
**Solution **

It may be noted that
*m*
*N*
*T*
*R*
*n*
*V*
*p*
*V*
*p _{i}*

*=*

_{i}

_{f}*= =40 •*

_{f}which indicates that the given process is isothermal. The work done at constant temperature is given by

*J*
*n*
*p*
*p*
*n*
*T*
*R*
*n*
*W*
*f*
*i*
5
.
55
)
40
/
10
(
)
40
( =−
=
= λ λ
**Problem 3-21 **

**A balloon contains ** 0.30 *mol***of helium. It rises, while **
**maintaining at constant **300 *K***temperature, to an altitude **
**where its volume has expanded five times. How much work **
**is done by the gas in the balloon during this isothermal **
**expansion? **

**Solution **

The work done at constant temperature is given by

_{}
=
*i*
*f*
*V*
*V*
*n*
*T*
*R*
*n*
*W* λ
*J*
*V*
*V*
*n*
*W* 3
0
0
10
203
.
1
5
)
300
)(
314
.
8
)(
30
.
0
( _{}= ×
= λ
**Problem 3-22 **

**One mole of nitrogen gas is compressed isothermally from **

2 /

10 *N* *m* ** to ** 2

/

20 *N* *m* ** at **270*C***. Calculate the work done. **
**Solution **

The work done during this compression is given by

_{}
−
=
*i*
*f*
*p*
*p*
*n*
*T*
*R*
*n*
*W* λ
*J*
*n*
*W* 1729
10
20
)
273
27
)(
314
.
8
)(
1
( =−
+
−
= λ

**3-2(D) WORK DONE IN THERMAL ISOLATION **
**OR UNDER ADIABATIC CONDITIONS **
**Problem 3-23 **

**One mole of oxygen, initially kept at **170*C***, is adiabatically **
**compressed so that its pressure becomes ten times. **
**Calculate **

**(a) its temperature after the compression and **
**(b) the work done on the gas. **

**Given that ***Cv* =21.1 *J* *mol*−1*K*−1** is for oxygen. **
**Solution **

(a) For adiabatic process

*p _{f}V_{f}*γ =

*p*γ

_{i}V_{i}*p*

_{f}*p*−γ (

_{f}*p*)γ =

_{f}V_{f}*p*−γ (

_{i}p_{i}*p*)γ γ γ γ γ ) ( ) ( 1 1

_{i}V_{i}*i*

*i*

*f*

*f*

*n*

*R*

*T*

*p*

*n*

*R*

*T*

*p*− = −

*p*1

*−γ*

_{f}*T*γ =

_{f}*p*1−γ

_{i}*T*γ γ γ)/ 1 (− =

_{i}*f*

*i*

*i*

*f*

*p*

*p*

*T*

*T*Now

*T*=170

_{i}*C*=(17+273)

*K*=290

*K*0.1 10 1 = =

*f*

*i*

*p*

*p*, γ =1.40 and 7 2 40 . 1 40 . 1 1 1 − = − = − γ γ Hence

*T*=(290)(0.1)−2/7 =560

_{f}*K*(b)

*dW*=

*dE*=

_{INTERNAL}*m*

*C*

_{v}dT*J*

*dW*=(1)(21.1)(560−290)=5697

**Problem 3-24**

**Calculate the work done to compress adiabatically one gram **
**mole of air initially at S.T.P. conditions to half its volume if **

40
.
1
=
γ ** for air. **
**Solution **

The initial volume *V*_{0}of air at S.T.P. is given by
*p*_{1}*V*_{1} =*n* *R* *T*_{1}
3
1
1
1
*p*
*T*
*R*
*n*
*V* =
*m*
*V*
*V*_{1} _{0} _{5} 2.24 10 2
10
013
.
1
)
273
0
)(
314
.
8
)(
1
( _{=} _{×} −
×
+
=
=

The final pressure of air, under adiabatic conditions, is given by
γ
=
*f*
*i*
*i*
*f*
*V*
*V*
*p*
*p*
*Pa*
*V*
*V*
*p _{f}* 5
40
.
1
0
0
5
10
673
.
2
5
.
0
)
10
013
.
1
(

_{}= × × =

The desired work done is

1
−
−
=
γ
*i*
*i*
*f*
*fV* *pV*
*p*
*W*
40
.
1
1
)
10
24
.
2
)(
10
013
.
1
(
)}
10
24
.
2
(
)
5
.
0
){(
10
673
.
2
( 5 2 5 2
−
×
×
−
×
×
×
=
−
−
*W*
*J*
*W* =−1812
**Problem 3-25 **
3

*2 m* **of a gas at ** 100 *N*/*m*2**expands according to law **
*C*

*V*
*P* 1.2 =

**where C is a constant, until volume is doubled. **
**Calculate the work done. **

**Solution **
Now *P _{i}V_{i}*1.2 =

*C*7 . 229 ) 2 )( 100 ( 1.2 = =

*C*Hence

*P*

*V*1.2 =229.7

*V*

*p*=229.7

*W* *p* *dV* *V* *dV*
*V*
*V*

### ∫

_{=}

### ∫

− = 2 1 4 2 2 . 1 7 . 229## [

## ]

*J*

*V*

*W*4 2 129.43 2 . 0 7 . 229 2 . 0 7 . 229 0.2 0.2 4 2 2 . 0 = − − = − = − − −

**Problem 3-26**

**Five litres of argon at ** 00*C*** are compressed to one litre **
**adiabatically and reversibly. What will be the final **
**temperature if **γ =5/3** for argon? **

**Solution **

For an adiabatic process
*p _{f}V_{f}*γ =

*p*γ γ

_{i}V_{i}

_{i}_{1}γ

*i*

*i*

*f*

*f*

*f*

*V*

*V*

*T*

*R*

*n*

*V*

*V*

*T*

*R*

*n* = Θ

*p*

*V*=

*n*

*R*

*T*

*T*γ−1 =

_{f}V_{f}*T*γ−1 1 − = γ

_{i}V_{i}*f*

*i*

*i*

*f*

*V*

*V*

*T*

*T*

*K*

*T*=(273)(5/1)(5/3)−1=798

_{f}**Problem 3-27**

**Calculate the final temperature of a sample of carbon **
**dioxide of mass ** 16.0 *g***that is expanded reversibly and **
**adiabatically from ***500 cm*3**at **298.15 *K***to***2000 cm*3**. **

**Solution **

For an adiabatic process
1
−
=
γ
*f*
*i*
*i*
*f*
*V*
*V*
*T*
*T*
*K*
*T _{f}* 196.71
2000
500
)
15
.
298
(
1
30
.
1
=
=
−

**Problem 3-28**

**By how much must the volume of a gas with **γ =1.40** be **
**changed to an adiabatic process if the Kelvin temperature is **
**to double? **

**Solution **

For an adiabatic process
*p _{f}V_{f}*γ =

*p*γ γ

_{i}V_{i}

_{i}_{1}γ

*i*

*i*

*f*

*f*

*f*

*V*

*V*

*T*

*R*

*n*

*V*

*V*

*T*

*R*

*n* = Θ

*p*

*V*=

*n*

*R*

*T*

*T*γ−1 =

_{f}V_{f}*T*γ−1

_{i}V_{i}*f*

*i*

*i*

*f*

*T*

*T*

*V*

*V*= γ−1 ) 1 /( 1 − = γ

*f*

*i*

*i*

*f*

*T*

*T*

*V*

*V*0.177 2 ) 1 40 . 1 /( 1 = = −

*T*

*T*

*V*

*V*

*i*

*f*

*V*=0.177

_{f}*V*

_{i}**Problem 3-29**

**A **1.00 *mol*** sample of an ideal diatomic gas originally at **
*atm*

00 .

1 ** and ** 200*C***, expands adiabatically to twice its **
**volume. What are final pressure and temperature for the **
**gas? Assume no molecular vibrations. **

**Solution **

For an adiabatic process
*p _{f}V_{f}*γ =

*p*γ γ =

_{i}V_{i}*f*

*i*

*i*

*f*

*V*

*V*

*p*

*p*

40
.
1
0
0
2
)
1
( _{}
=
*V*
*V*

*pf* Θ γ =1.40 for a diatomic gas.

*p _{f}* =0.379

*atm*Further

*T*γ−1 =

_{f}V_{f}*T*γ−1 1 40 . 1 0 0 1 2 ) 273 20 ( − − + = =

_{i}V_{i}*V*

*V*

*V*

*V*

*T*

*T*

*f*

*i*

*i*

*f*γ

*T*=222

_{f}*K*

**Problem 3-30**

**An ideal gas initially at **8.00 *atm***and **300 *K***is permitted to **
**expand adiabatically until its volume doubles. Find the final **
**pressure and temperature if the gas is **

**(a) Monoatomic (b) Diatomic **
**Solution **

The expressions for final pressure and temperature of the gas
are given by
γ
=
*f*
*i*
*i*
*f*
*V*
*V*
*p*
*p*
and
1
−
=
γ
*f*
*i*
*i*
*f*
*V*
*V*
*T*
*T*

(a) For monoatomic gas

3
5
=
γ therefore
*atm*
*V*
*V*
*p*
*i*
*i*
*f* 2.52
2
)
00
.
8
(
3
/
5
=
=
*K*
*V*
*V*
*T*
*i*
*i*
*f* 189
2
)
300
(
1
)
3
/
5
(
=
=
−

(b) For diatomic gas 1.40
5
7
=
=
γ therefore
*atm*
*V*
*V*
*p*
*i*
*i*
*f* 3.03
2
)
00
.
8
(
40
.
1
=
=
*K*
*V*
*V*
*T*
*i*
*i*
*f* 227
2
)
300
(
1
40
.
1
=
=
−
**Problem 3-31 **

**A volume of dry air at S.T.P. is expanded to three times its **
**original volume under adiabatic conditions. Calculate the **
**final temperature and pressure if **γ =1.40** for air. **

**Solution **

The final temperature of air is given by
1
−
=
γ
*f*
*i*
*i*
*f*
*V*
*V*
*T*
*T*
*K*
*V*
*V*
*T _{f}* 176
3
)
273
0
(
40
.
1
0
0

_{=} + =

The final pressure of the gas is given by
γ
=
*f*
*i*
*i*
*f*
*V*
*V*
*p*
*p*
*Pa*
*V*
*V*
*pf*
4
40
.
1
0
0
5
10
176
.
2
3
)
10
013
.
1
( _{} = ×
×
=
**Problem 3-32 **

**An ideal monoatomic gas for which **γ =5/3** undergoes an **
**adiabatic expansion to one third of its initial pressure. Find **
**the ratio of final volume to initial volume if the process is **
** (a) Isothermal **

**Solution **

(a) For an isothermal process
*p _{f}V_{f}* =

*p*3 ) 3 / 1 ( = = =

_{i}V_{i}*i*

*i*

*f*

*i*

*i*

*f*

*p*

*p*

*p*

*p*

*V*

*V*

(b) For an adiabatic process
*p _{f}V_{f}*γ =

*p*γ

_{i}V_{i}*f*

*i*

*i*

*f*

*p*

*p*

*V*

*V*= γ 933 . 1 ) 3 ( ) 3 / 1 ( 6 . 0 ) 3 / 5 /( 1 / 1 = = = =

*i*

*i*

*f*

*i*

*i*

*f*

*p*

*p*

*p*

*p*

*V*

*V*γ

**Example 3-33**

**A volume of argon gas at **270*C***expands adiabatically until **
**its volume is increased four times. Find the resulting fall in **
**temperature. Given that**γ =1.67**is for argon gas. **

**Solution **
Now
1
−
=
γ
*f*
*i*
*i*
*f*
*V*
*V*
*T*
*T*
*C*
*or*
*K*
*V*
*V*
*T _{f}* 0
1
67
.
1
0
0

_{118}

_{.}

_{5}

_{154}

_{.}

_{5}4 ) 273 27 (

_{}= − + = −

**Example 3-34**

**An ideal gas at **300 *K*** is compressed adiabatically to half **
**its initial volume. **

**(a) What is the final temperature of the gas if it is **
**monoatomic? **

**Solution **
Now
1
−
=
γ
*f*
*i*
*i*
*f*
*V*
*V*
*T*
*T*

(a) For monoatomic gas

3
5
=
γ therefore
*K*
*V*
*V*
*T _{f}* 476
5
.
0
)
300
(
1
)
3
/
5
(
0
0

_{=} = −

(b) For diatomic gas 1.40
5
7
=
=
γ therefore
*K*
*V*
*V*
*T _{f}* 396
5
.
0
)
300
(
1
40
.
1
0
0

_{=} = −

**Example 3-35**

**Calculate the rise in temperature when a gas at ** 270*C***is **
**compressed to eight times its original pressure. The value of **

**γ is 1.5 for the given gas. **
**Solution **

The final temperature of the gas is given by the expression
γ
γ)/
1
(−
=
*f*
*i*
*i*
*f*
*p*
*p*
*T*
*T*
*K*
*p*
*p*
*T*
*i*
*i*
*f* 600
8
)
273
27
(
5
.
1
/
)
5
.
1
1
(
=
+
=
−

The rise in temperature of the gas will be

*K*
*T*
*T*
*T* = *f* − *i* =600−300=300
∆
**Example 3-36 **

**A given mass of gas at ** 00*C***is suddenly compressed to a **
**pressure twenty times the initial pressure. What will be the **
**final temperature of the gas if γ is **1.42** . **

**Solution **

The final temperature of the gas is given by the expression
γ
γ)/
1
(−
=
*f*
*i*
*i*
*f*
*p*
*p*
*T*
*T*
*K*
*p*
*p*
*T*
*i*
*i*
*f* 662.2
20
)
273
0
(
42
.
1
/
)
42
.
1
1
(
=
+
=
−
**Example 3-37 **

**One mol of an ideal monoatomic gas at **300 *K***and **3.0 *atm*
**expands adiabatically to a final pressure of**1.0 *atm***. How **
**much work does the gas do in the expansion? **

**Solution **

For adiabatic process
γ
γ)/
1
(−
=
*f*
*i*
*i*
*f*
*p*
*p*
*T*
*T*
*K*
*T _{f}* 193
0
.
1
0
.
3
)
300
(
67
.
1
/
)
67
.
1
1
(
=
=
−

The work done in adiabatic process is given by

*J*
*T*
*T*
*R*
*W* _{i}* _{f}* 1328
1
67
.
1
)
193
300
(
314
.
8
)
(
1 − =
−
=
−
−
=
γ

**Example 3-38**

**An ideal monoatomic gas, consisting of ** 2.6 *mol***of volume **

3 084 .

0 *m* **, expands adiabatically. The initial and final **
**temperatures are ** 250*C*** and ** −680*C***respectively. What is **
**the final volume of the gas? **

**Solution **

For adiabatic process

1
1 −
−
= γ
γ
*i*
*f*
*i*
*f* *V*
*T*
*T*
*V*
)
1
/(
1 −
=
γ
*f*
*i*
*f*
*T*
*T*
*V*
Now
** ** 3
084
.
0 *m*
*Vi* =
*T _{i}* =250

*C*=(25+273)

*K*=298

*K*

*K*

*K*

*C*

*T*=−680 =(−68+273) =205

_{f} γ =1.67 for monoatomic gas.
Hence
3
)
1
67
.
1
/(
1
2 (0.084) 0.147
205
298
*m*
*V* =
=
−

**ADDITIONAL PROBLEMS **

**(1) A system receives **150**calories of heat and the change **
**in its internal energy is ** 209**joules. Calculate the **
**work done by the system. (**1 *calorie*=4.18 *J***) **

** B.U. B.Sc. 2007A **
**(2) The atmospheric pressure is ** 6 2

/ 10

013 .

1 × *dynes* *cm* **. A **

**gas expands at this pressure and the increase in its **

**volume is **668 *c*.*c*.**Find the work done by the gas. **
** B.U. B.Sc. (Hons.) 1989A **

**(3) A gas expands at atmospheric pressure and its **
**volume increases by***400 cm*3**. Find the work done by **
**the gas. B.U. B.Sc. (Hons.) 1991A **
**(4) A gas is suddenly compressed to one fourth of its **

**original volume. Calculate rise in temperature, the **
**original at **270*C***and **γ =1.5**. F.P.S.C. 1978 **

**Answers **

(1) 418 J (2) 67.67 J (3) 40.52 J (4) 600 K