If the volume of $N_2$ gas at $STP$ is $204.75 \, mL$,then calculate the volume of the gas at $1.5 \, bar$ pressure and $127 \, ^oC$ temperature.
(200 ML) Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
At $STP$,$T_1 = 273 \, K$ and $P_1 = 1 \, bar$.
Given $V_1 = 204.75 \, mL$.
For the final state: $T_2 = 127 + 273 = 400 \, K$ and $P_2 = 1.5 \, bar$.
Substituting the values: $\frac{1 \times 204.75}{273} = \frac{1.5 \times V_2}{400}$.
Solving for $V_2$: $V_2 = \frac{204.75 \times 400}{273 \times 1.5} = 200 \, mL$.
At $STP$,$T_1 = 273 \, K$ and $P_1 = 1 \, bar$.
Given $V_1 = 204.75 \, mL$.
For the final state: $T_2 = 127 + 273 = 400 \, K$ and $P_2 = 1.5 \, bar$.
Substituting the values: $\frac{1 \times 204.75}{273} = \frac{1.5 \times V_2}{400}$.
Solving for $V_2$: $V_2 = \frac{204.75 \times 400}{273 \times 1.5} = 200 \, mL$.
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