At 400 , K temperature,200 , mL of N_2 gas ha | vedclass

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(A) Given: $P_1 = 1.5 \, bar$,$V_1 = 200 \, mL$,$T_1 = 400 \, K$. At $STP$ (Standard Temperature and Pressure),$P_2 = 1 \, bar$ and $T_2 = 273.15 \, K

Vedclass · Accepted Answer

$204.75$

Vedclass · Answer

(A) Given: $P_1 = 1.5 \, bar$,$V_1 = 200 \, mL$,$T_1 = 400 \, K$. At $STP$ (Standard Temperature and Pressure),$P_2 = 1 \, bar$ and $T_2 = 273.15 \, K$. Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$. Substituting the values: $\frac{1.5 	imes 200}{400} = \frac{1 	imes V_2}{273.15}$. $0.75 = \frac{V_2}{273.15}$. $V_2 = 0.75 	imes 273.15 = 204.86 \, mL$ (approximately $204.75 \, mL$ depending on rounding).

At $400 \, K$ temperature,$200 \, mL$ of $N_2$ gas has a pressure of $1.5 \, bar$. Find the volume of $N_2$ gas at $STP$. (in $, mL$)

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