Question

One mole of an ideal monatomic gas undergoes the following four reversible processes:
$Step1$ - it is first compressed adiabatically from volume $8.0m^3$ to $1.0m^3$.
$Step2$ - then expanded isothermally at temperature $T_1$ to volume $10.0m^3$.
$Step3$ - then expanded adiabatically to volume $80.0m^3$.
$Step4$ - then compressed isothermally at temperature $T_2$ to volume $8.0m^3$.
Then $T_1/T_2$ is

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Answer: (B)

$4$

Given:

$V_1=8m^3$

$V_2=1m^3$

$V_3=10m^3$

$V_4=80m^3$

To find $\frac{T_1}{T_2}=?$

for ourcase $\to \frac{T_2}{T_4}=?$ (as per our diagram)

Process: $\{2\to 3\&4\to 1\}\to$ Isothermal process

$T_2=T_3\&T_4=T_1$

Process $1\to 2$

we know for adiabatic process $\to \frac{T_1}{T_2}={\left(\frac{V_2}{V_1}\right)}^{\gamma -1}\Rightarrow \frac{T_1}{T_2}={\left(\frac{1}{8}\right)}^{\gamma -1}$ ...(i)

$3\to 4$ ad. expn.

$\frac{T_4}{T_3}={\left(\frac{V_3}{V_4}\right)}^{\gamma -1}={\left(\frac{1}{8}\right)}^{\gamma -1}$ ...(ii)

from eqn. (i) & (ii)

$\gamma =1.66$ $\frac{T_1}{T_2}=\frac{T_4}{T_3}\Rightarrow \frac{T_4}{T_2}={\left(\frac{1}{8}\right)}^{\gamma -1}\Rightarrow \frac{T_4}{T_2}=\frac{1}{4}$

( Since $T_1=T_4\&[\frac{T_4}{T_3}={\left(\frac{1}{8}\right)}^{\gamma -1}]$)

$\frac{T_2}{T_4}=2$

That is $\frac{T_1^{\prime }}{T_2}=4$