Question

Helium at ${27^0}$C has a volume of 8 liters. It is suddenly compressed to a volume of 1 liter. The temperature of the gas will be$\gamma = \dfrac{5}{3}$
A. ${108^0}$C
B. ${9327^0}C$
C. ${1200^0}C$
D. ${927^0}C$

Answer 1

Hint: In order to solve this numerical we should know about how the volume and temperature of an ideal gas vary during a reversible adiabatic compression or expansion. By using the equation of the same we can solve the problem.

Complete step by step answer:
From the given data:
${T_1} = {27^0}C$
$\gamma = \dfrac{5}{3}$ Where $\gamma $ is the ratio of specific heat capacity at constant pressure to the specific heat capacity at constant Volume.
By using the equation of temperature and volume of an ideal gas vary during a reversible adiabatic expansion or compression is
${T_1}{V_1}^{\gamma - 1} = {T_2}{V_2}^{\gamma - 1}$
${T_2} = {\left( {\dfrac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}}{T_1}$
Substituting the $\gamma $ value to the above equation we get
${T_2} = {\left( {\dfrac{{{V_1}}}{{{V_2}}}} \right)^{\dfrac{5}{3} - 1}} \times 300$
$\
\implies {T_2} = {\left( {\dfrac{8}{1}} \right)^{\dfrac{5}{3} - 1}} \times 300 \\
\implies {T_2} = {\left( {\dfrac{8}{1}} \right)^{\dfrac{2}{3} - 1}} \\
\ $
$\
\implies {T_2} = 4 \times 300 \\
\implies {T_2} = 1200K \\
\ $
$\
\implies {T_2} = 1200 - 273 \\
\therefore {T_2} = {927^0}C \\
\ $

So, the correct answer is “Option D”.

Note:
Students should understand the process of adiabatic expansion and adiabatic compression. Adiabatic expansion is the ideal behavior for a closed system, temperature is decreasing at constant pressure. Adiabatic compression is the process in which no heat is exchanged and increases in internal energy which is equal to the external work done.