An ideal gas at a pressure of 1 text{ atm} an | vedclass

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(D) For an adiabatic process,the relationship between temperature and pressure is given by $\frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\g

Vedclass · Accepted Answer

$327$

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(D) For an adiabatic process,the relationship between temperature and pressure is given by $\frac{T_2}{T_1} = \left( \frac{P_2}{P_1} ight)^{\frac{\gamma - 1}{\gamma}}$.Given: $P_1 = 1 	ext{ atm}$,$P_2 = 8 	ext{ atm}$,$T_1 = 27^{\circ}C = 300 	ext{ K}$,and $\gamma = 3/2$.Substituting the values: $\frac{T_2}{300} = (8)^{\frac{3/2 - 1}{3/2}} = (8)^{\frac{1/2}{3/2}} = (8)^{1/3}$.Since $8^{1/3} = 2$,we have $\frac{T_2}{300} = 2$.Therefore,$T_2 = 600 	ext{ K}$.Converting back to Celsius: $T_2(^{\circ}C) = 600 - 273 = 327^{\circ}C$.

An ideal gas at a pressure of $1 \text{ atm}$ and temperature of $27^{\circ}C$ is compressed adiabatically until its pressure becomes $8$ times the initial pressure. The final temperature is ..... $^{\circ}C$ (given $\gamma = 3/2$).

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