Question

What is the final temperature of 0.10 mole monatomic ideal gas that performs 75 cal of work adiabatically if the initial temperature is 227 degree celsius?

Answer 1

Let the final Temperature be T. (gamma=5/3)Now we can apply TV^(gamma-1)=constant.300(16)^2/3=T(2)^2/3(8)^2/3=T/3004=T/300We have, T =1200K

Answer 2

Answer : The final temperature is, $447^oC$

Explanation : Given,

Moles of gas = 0.10 mole

Work done = 75 cal

Initial temperature = $227^oC$

The used for work done in reversible adiabatic expansion for an ideal gas will be,

$w=\frac{1}{(\gamma-1)}\times nR\times (T_2-T_1)$

where,

w = work done

$\gamma$ = for monoatomic gas = $\frac{5}{3}$

n = number of moles

R = gas constant = 2 cal

$T_2$ = final temperature = ?

$T_2$ = initial temperature

Now put all the given values in the above formula, we get the final temperature of the gas.

$75cal=\frac{1}{(\frac{5}{3}-1)}\times (0.1mole)\times (2cal)\times (T_2-227^oC)$

$T_2=477^oC$

Therefore, the final temperature is, $447^oC$