Question

17. An ideal monatomic gas at 300 K expands
adiabatically to 8 times its volume. What is the final
temperature?
(1) 75 K
(2) 300 K
(3) 560 K
(4) 340 K​

Answer 1

Answer:

1.75 k

Explanation:

Solution:- (A) 75K

For an adiabatic process,

TV

γ−1

=constant

∴

T 1T

2

=(

V

2

V

1

)

γ−1

Given:-

T

1

=300K

T

2

=T(say)=?

V

1

=V

V

2

=8V

γ=

3

5

(∵the gas is monoatomic)

∴ 300

T=( 8V ) 35 −1

⇒ 300T

=(81) 32 ⇒T= 4300

=75K

Hence the final temperature will be 75K.

akka answer correct - ஆ

Answer 2

Answer:

Since It Is an Ideal Monoatomic Gas It's Degree of Freedom(f) Will be 3

and Let x be the Gammma

So, We Know

[ x = 1 + 2/f ]

Putting f = 3, We get

x = 5/3

• Now Since We Have Given The Initial Temperature($T_{1}$) as 300K

• Let Initial Volume($V_{1}$) be V and  We Have Given Final Volume($V_{2}$)is 8V

Now Since The Gas is expanding adiabatically (ΔQ = 0)

So We Can Use The Equation

$T_{1}V_{1}^{x-1} = T_{2}V_{2}^{x-1}$

[T2 is the Final Temperature]

Putting The Values

$300*V_{1}^{x-1} = T_{2}(8V_{2})^{x-1}$

$300 = T_{2}*8^{2/3}\\\\300 = T_{2} *4\\\\T_{2} = 75K$

So We Have Option(1) Correct Final Temperature is 75K