Question

If $2$ moles of an ideal gas at $546K$ occupy volume $44.8L$, then pressure must be :

• A. $2atm$
• B. $3atm$
• C. $4atm$
• D. $1atm$

Answer 1

The correct option is A $2atm$

From gas equation, $PV=nRT$

Given that :

$V=44.8L$
$n=2$
$R=0.0821$ $L$ $atm/mol/K$
$T=546K$

$P=\frac{nRT}{V}$

$=\frac{2\times 0.0821\times 546}{44.8}$

$=2atm$

Therefore, the option is A.