If the initial volume of a neon gas is reduce | vedclass

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(A) Using the ideal gas law equation $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.Given: $P_1 = P$,$P_2 = 2P$,$V_1 = V$,$V_2 = 0.85 V$ (since volume re

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$319$

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(A) Using the ideal gas law equation $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.Given: $P_1 = P$,$P_2 = 2P$,$V_1 = V$,$V_2 = 0.85 V$ (since volume reduced by $15\%$).Initial temperature $T_1 = 75 + 273.15 = 348.15 \, K$.Substituting the values: $\frac{P 	imes V}{348.15} = \frac{2P 	imes 0.85 V}{T_2}$.$T_2 = \frac{2 	imes 0.85 	imes 348.15}{1} = 591.855 \, K$.Converting back to Celsius: $T_2(^\circ C) = 591.855 - 273.15 = 318.705 \, ^\circ C \approx 319 \, ^\circ C$.

If the initial volume of a neon gas is reduced by $15\%$ upon cooling to $75\,^\circ C$,to what temperature in $^\circ C$ should it be heated so that the pressure becomes double?

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