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

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 $216K$.