Question

What is the final temperature of $0.10$ mole monatomic ideal gas performs $75cal$ of work adiabatically if the initial temperature is ${227}^{o}C$?
[Use : $R=2cal/(K-mol$)]

• A. $250K$
• B. $300K$
• C. $350K$
• D. $750K$

Answer 1

Answer: (D)

$750K$

Solution:- (D) $750K$

Let the final temperature be T.

Initial temperature $=227℃=(273+227)=500K$

For an adiabatic process, work done is given by-

$W=n{C}_{V}dT$

As the gas is monoatomic,

$C_V=\frac{3}{2}R$

$\therefore W=\frac{3}{2}nRdT$

Given that,

$n=$ No. of moles $=0.10$

$R=2cal{K}^{-1}{mol}^{-1}$

$W=$ Work done $=75cal$

Since the work done is positive, i.e., work is done by the system, means the gas has expanded.

$\therefore$ Change in temperature $\left(dT\right)=T_2-T_1=(T-500)K$

$\therefore$ From equation $\left(1\right)$, we have

$75=\frac{3}{2}\times 0.10\times 2\times (T-500)$

$\Rightarrow T-500=\frac{75}{0.3}$

$\Rightarrow T=500+250=750K$

Hence the final temperature is $750K$.