When 1 , text{mole} of monoatomic ideal gas a | vedclass

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

Vedclass · Accepted Answer

$T - \frac{2}{3 	imes 0.0821}$

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(A) For an adiabatic process,the first law of thermodynamics states that $\Delta U = q + w$. Since the process is adiabatic,$q = 0$,so $\Delta U = w$.For an ideal gas,$\Delta U = n C_v \Delta T$. For a monoatomic gas,$C_v = \frac{3}{2} R$.Thus,$n 	imes \frac{3}{2} R 	imes (T_f - T) = -P_{ext} 	imes \Delta V$.Given $n = 1 \, 	ext{mole}$,$P_{ext} = 1 \, 	ext{atm}$,$\Delta V = (2 - 1) \, 	ext{L} = 1 \, 	ext{L}$,and $R = 0.0821 \, 	ext{L atm K}^{-1} 	ext{mol}^{-1}$.Substituting the values: $1 	imes \frac{3}{2} 	imes 0.0821 	imes (T_f - T) = -1 	imes 1$.$(T_f - T) = -\frac{2}{3 	imes 0.0821}$.$T_f = T - \frac{2}{3 	imes 0.0821}$.

When $1 \, \text{mole}$ of monoatomic ideal gas at $T \, \text{K}$ undergoes adiabatic change under a constant external pressure of $1 \, \text{atm}$ and changes volume from $1 \, \text{L}$ to $2 \, \text{L}$,the final temperature in Kelvin would be:

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