Question

When 1 mole of a mono atomic ideal gas a T K undergoes an adiabatic change under a constant external pressure of 1 atm, its volume changes from 1L to 2L. The final temperature in Kelvin is:

• A. $\frac{T}{2^{2/3}}$
• B. $T+\frac{2}{3\times 0.0821}$
• C. $T$
• D. $T-\frac{2}{3\times 0.0821}$

Answer 1

The correct option is D $T-\frac{2}{3\times 0.0821}$

In an adiabatic process, $Q$ = 0
From first law of thermodynamics,
$\Delta U=Q+W$
$\Delta U=W$...1
Now $W$ = $-P(V_2-V_1$) (given $P_{ext}=1atm$
= -1(2-1) atm L = -1 atm L
For calculating $\Delta U$ we need to calculate $C_v$
$C_v$ = $\frac{R}{\gamma -1}$
for monoatomic gas $\gamma =\frac{5}{3}$
Hence $C_v=\frac{3R}{2}$
Now using eq. 1
$\Delta U=n{C}_{v}dT$
Now using eq. 1
$1\times \frac{3R}{2}(T_f-T_i)=-1atmL$
Assuming final temperature as $T_f$ and solving for it we get
$T_f=T_i-\frac{2}{3R}$