Question

What is the fall in temperature of helium initially at \[{15^0}C\]when at is suddenly expanded to \[8\]times its original volume \[\left( {\gamma = \dfrac{5}{3}} \right)\]
A. \[{203.01^0}C\]
B. \[{200.01^0}C\]
C. \[{216.01^0}C\]
D. None of these

Answer 1

Hint: Helium is a gas when it is expanded suddenly it can be called an adiabatic expansion. The temperature and volume are directly proportional to each other. The final volume will be \[8\]times its original volume. The isentropic factor is given and the final temperature can be calculated.
Formula used:
\[{T_1}{V_1}^{\gamma - 1} = {T_2}{V_2}^{\gamma - 1}\]
Where \[{T_1}\] is initial temperature
\[{T_2}\]is final temperature
\[{V_1}\] is initial volume be \[V\]
\[{V_2}\] is final volume be \[8V\]
isentropic factor \[\gamma = \dfrac{5}{3}\]

Complete answer:
Helium is a gas and adiabatic expansion means the gas is suddenly expanded.
Given that the initial temperature will be \[{15^0}C\], it should be converted into kelvins.
The temperature in kelvin is \[288K\]
Let the initial volume will be V
The final volume is suddenly expanded to \[8\]times its original volume.
Given isentropic factor \[\gamma = \dfrac{5}{3}\]
The final temperature can be calculated from the above formula
By substituting all the above values, we will get
\[{T_2} = 288{\left( {\dfrac{V}{{8V}}} \right)^{\dfrac{5}{3} - 1}}\]
By simplifying we will get \[{T_2} = 72K\]
The initial temperature is \[288K\] and final temperature is \[72K\]
Thus, the difference gives the fall in temperature \[288 - 72 = 216K\]
This temperature will be equal to \[{216^0}C\]

So, the correct answer is “Option C”.

Note:
Though the values of kelvin and Celsius were different in the values of temperature, the magnitude of those two were same. There are both the units of temperature. Thus, a degree Celsius will be equal to one degree kelvin. In the above problem \[{216^0}C\] will be equal to \[216K\].