At what ^o C temperature,the sample of neon g | vedclass

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(A) Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$ Initial conditions: $P_1 = P$,$V_1 = V$,$T_1 = 75 + 273 = 348 \ K$ Final c

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$319$

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(A) Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$ Initial conditions: $P_1 = P$,$V_1 = V$,$T_1 = 75 + 273 = 348 \ K$ Final conditions: $P_2 = 2P$,$V_2 = 0.15V$ (since volume is reduced to $15\%$ of its initial value) Substituting the values: $\frac{P 	imes V}{348} = \frac{2P 	imes 0.15V}{T_2}$ $T_2 = 348 	imes 2 	imes 0.15 = 104.4 \ K$ Converting to Celsius: $T_2(^oC) = 104.4 - 273 = -168.6 \ ^oC$ Note: Based on the calculation,the provided options do not match the result. However,if the question meant volume is reduced $BY$ $15\%$ (i.e.,$V_2 = 0.85V$),then $T_2 = 348 	imes 2 	imes 0.85 = 591.6 \ K = 318.6 \ ^oC$,which rounds to $319 \ ^oC$ (Option $A$).

At what $^o C$ temperature,the sample of neon gas would be heated to double its pressure,if the initial volume of gas is reduced to $15\%$ at ${75\,^o}C$?

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