Question 426: Thermodynamics & KTG

Question

Two moles of an ideal monatomic gas occupies a volume V at 27
o
C. The gas expands adiabatically to a
volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

Accepted Answer

For adiabatic process,
pv
$$\gamma $$
= constant.
and we know, pv = nRT
$$	herefore\,\,\,$$ p = $${{nRT} \over v}$$
$$	herefore\,\,\,$$ $${{nRT} \over v} 	imes {v^\gamma }$$ = constant
$$ \Rightarrow $$$$\,\,\,$$ T v
$$\gamma $$$$-$$1
= constant.
$$	herefore\,\,\,$$ T
1
v
1
$$\gamma $$$$-$$1
= T
2
v
2
$$\gamma $$$$-$$1
Given that,
This is a monoatomic gas.
So, degree of frequency f = 3
$$	herefore\,\,\,$$ $$\gamma $$ = $${{{c_p}} \over {{c_v}}}$$ = 1 + $${2 \over f}$$ = 1 + $${2 \over 3}$$ = $${5 \over 3}$$
T
1
= 27 + 273 = 300 K
v
1
= v
and v
2
= 2v
$$	herefore\,\,\,$$ T
2
(2V)$$^{{2 \over 3}}$$ = 300(v)$$^{{2 \over 3}}$$
$$ \Rightarrow $$$$\,\,\,$$ T
2
= $${{300} \over {{2^{{2 \over 3}}}}}$$ = 189 K
Change in internal energy,
$$\Delta $$U = $${1 \over 2}$$ nfR$$\Delta $$T
= $${1 \over 2}$$ $$ 	imes $$ 2 $$ 	imes $$ 3 $$ 	imes $$ 8.31 $$ 	imes $$ (189 $$-$$ 300)
= $$-$$ 2.7 KJ

Answer

(a) 195 K  (b) 2.7 kJ

Answer

(a) 189 K (b) 2.7 kJ

Answer

(a) 195 K (b) –2.7 kJ

Answer

(a) 189 K (b) – 2.7 kJ

Thermodynamics & KTG

Two moles of an ideal monatomic gas occupies a volume V at 27 o C. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

Choose the correct answer: