At 0, ^circ C and 1, text{atm} pressure,a gas | vedclass

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(B) Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$Given:$P_1 = P$$V_1 = 100\, 	ext{cc}$$T_1 = 0\, ^\circ C = 273\, 	ext{K}$

Vedclass · Accepted Answer

$88.9$

Vedclass · Answer

(B) Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$Given:$P_1 = P$$V_1 = 100\, 	ext{cc}$$T_1 = 0\, ^\circ C = 273\, 	ext{K}$New conditions:$P_2 = 1.5 P = \frac{3}{2} P$$T_2 = T_1 + \frac{1}{3} T_1 = \frac{4}{3} T_1 = \frac{4}{3} 	imes 273\, 	ext{K}$Substituting these values into the formula:$V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{P 	imes 100 	imes (\frac{4}{3} 	imes 273)}{(\frac{3}{2} P) 	imes 273}$$V_2 = 100 	imes \frac{4}{3} 	imes \frac{2}{3} = \frac{800}{9} \approx 88.88\, 	ext{cc}$Rounding to one decimal place,$V_2 = 88.9\, 	ext{cc}$.

At $0\, ^\circ C$ and $1\, \text{atm}$ pressure,a gas occupies $100\, \text{cc}$. If the pressure is increased to $1.5$ times and the temperature is increased by one-third of its absolute temperature,then the final volume of the gas will be :- $cc$

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