Calculate the final temperature of a monoatom | vedclass

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(C) For a reversible adiabatic process,the relationship between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Given: $T

Vedclass · Accepted Answer

$1200$

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(C) For a reversible adiabatic process,the relationship between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Given: $T_1 = 300 \, K$,$V_1 = 16 \, L$,$V_2 = 2 \, L$.For a monoatomic ideal gas,the heat capacity ratio $\gamma = \frac{5}{3}$,so $\gamma - 1 = \frac{2}{3}$.Substituting the values: $300 	imes (16)^{2/3} = T_2 	imes (2)^{2/3}$.$T_2 = 300 	imes (\frac{16}{2})^{2/3} = 300 	imes (8)^{2/3}$.Since $(8)^{2/3} = (2^3)^{2/3} = 2^2 = 4$,we get $T_2 = 300 	imes 4 = 1200 \, K$.

Calculate the final temperature of a monoatomic ideal gas that is compressed reversibly and adiabatically from $16 \, L$ to $2 \, L$ at $300 \, K$. (in $, K$)

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