what is the temperature of 0.750 mol of a gas stored in a 6,850 ml cylinder at 2.21 atm? use $pv = nrt$ and…

what is the temperature of 0.750 mol of a gas stored in a 6,850 ml cylinder at 2.21 atm? use $pv = nrt$ and $r = 0.0821\frac{lcdot atm}{molcdot k}$. 2.95 k 5.24 k 138 k 246 k

what is the temperature of 0.750 mol of a gas stored in a 6,850 ml cylinder at 2.21 atm? use $pv = nrt$ and $r = 0.0821\frac{lcdot atm}{molcdot k}$. 2.95 k 5.24 k 138 k 246 k

Answer

Explanation:

Step1: Convert volume to liters

$V = 6850\ mL=6.85\ L$

Step2: Rearrange the ideal - gas law for temperature

$T=\frac{PV}{nR}$

Step3: Substitute the given values

$P = 2.21\ atm$, $n = 0.750\ mol$, $V = 6.85\ L$, $R=0.0821\frac{L\cdot atm}{mol\cdot K}$ $T=\frac{2.21\ atm\times6.85\ L}{0.750\ mol\times0.0821\frac{L\cdot atm}{mol\cdot K}}$ $T=\frac{15.1385}{0.061575}K\approx246\ K$

Answer:

D. 246 K