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# Calculate the temperature at which 28g $N_2$ occupies a volume of 10 litres at 2.46 atm.

$(a)\;399.64K\qquad(b)\;299.64K\qquad(c)\;199.64\qquad(d)\;236K$

W = 28g
P = 2.46 atm
V = 10 litres
m = 28
$\therefore PV = \large\frac{w}{m}$RT (R = 0.0821 litre atm/K/mol)
$T = \large\frac{mPV}{w\times R}$
$T = \large\frac{28\times2.46\times10}{28\times0.0821}$
$= \large\frac{688.8}{2.2988}$
= 299.64 K