What is the final temperature of 0.10 mol mo | vedclass

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(A) For an adiabatic process,the first law of thermodynamics is given by $\Delta U = q + w$. Since the process is adiabatic,$q = 0$,so $\Delta U = w$.

Vedclass · Accepted Answer

$250$

Vedclass · Answer

(A) For an adiabatic process,the first law of thermodynamics is given by $\Delta U = q + w$. Since the process is adiabatic,$q = 0$,so $\Delta U = w$.For an ideal gas,the change in internal energy is $\Delta U = nC_v \Delta T$.For a monoatomic gas,$C_v = \frac{3}{2}R$.Given: $n = 0.10 \ mol$,$w = -75 \ cal$ (work done by the system),$T_i = 227 + 273 = 500 \ K$,and $R = 2 \ cal \cdot K^{-1} \cdot mol^{-1}$.Substituting the values: $n 	imes \frac{3}{2} 	imes R 	imes (T_f - T_i) = w$.$0.10 	imes \frac{3}{2} 	imes 2 	imes (T_f - 500) = -75$.$0.3 	imes (T_f - 500) = -75$.$T_f - 500 = \frac{-75}{0.3} = -250$.$T_f = 500 - 250 = 250 \ K$.

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

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