Question

What is the final temperature of 0.10 mole
monoatomic ideal gas that performs 75 cal of
work adiabatically, if the initial temperature is
227°C (use R = 2 cal K-? mol-1)
(1) 100°C (2) 250°C (3) 150°C (4) 210°C​

Answer 1

Correct Question:-

What is the final temperature of $\sf{0.10\ mol}$  monoatomic ideal gas that performs $\sf{75\ cal}$ work adiabatically, if the initial temperature is  $\sf{227^oC\,?}$ (use $\sf{R=2\ cal\,K^{-1}\,mol^{-1})}$

$\sf{(1)\ 100\ K}$

$\sf{(2)\ 250\ K}$

$\sf{(3)\ 150\ K}$

$\sf{(4)\ 210\ K}$

Solution:-

Work done by an ideal gas in an adiabatic process is given by,

$\longrightarrow\sf{W=\dfrac{nR(T_2-T_1)}{1-\gamma}}$

where,

• $\sf{n=}$ no. of moles

• $\sf{R=2\ cal\,K^{-1}\,mol^{-1}}$

• $\sf{T_1=}$ initial temperature

• $\sf{T_2=}$ final temperature

• $\sf{\gamma=}$ heat capacity ratio, i.e., $\sf{\dfrac{C_P}{C_V}.}$

Here,

$\longrightarrow\sf{W=75\ cal}$

$\longrightarrow\sf{n=0.1\ mol}$

$\longrightarrow\sf{T_1=227^oC=500\ K}$

Since the ideal gas is monoatomic,

$\longrightarrow\sf{\gamma=\dfrac{5}{3}}$

Then,

$\longrightarrow\sf{W=\dfrac{nR(T_2-T_1)}{1-\gamma}}$

$\longrightarrow\sf{W(1-\gamma)=nR(T_2-T_1)}$

$\longrightarrow\sf{T_2-T_1=\dfrac{W(1-\gamma)}{nR}}$

$\longrightarrow\sf{T_2=T_1+\dfrac{W(1-\gamma)}{nR}}$

Substituting values for each in RHS,

$\longrightarrow\sf{T_2=500+\dfrac{75\left(1-\dfrac{5}{3}\right)}{0.1\times2}}$

$\longrightarrow\sf{\underline{\underline{T_2=250\ K}}}$

Hence the final temperature is $\bf{250\ K.}$