Using the ideal gas law PV =nRTPV=nRT , we find that the pressure will be P =\frac{nRT}{V}P=
V
nRT
. Then, we'll substitute and find the pressure, using T = -25 °C = 248.15 K and R = 0.0821 \frac{atm\cdot L}{mol \cdot K}
mol⋅K
atm⋅L
:
P =\frac{nRT}{V} = \frac{(0.33\,\cancel{mol})(0.0821\frac{atm\cdot \cancel{L}}{\cancel{mol \cdot K}})(248.15\,\cancel{K})}{15.0\,\cancel{L}} = 0.4482\,atmP=
V
nRT
=
15.0
L
(0.33
mol
)(0.0821
mol⋅K
atm⋅
L
)(248.15
K
)
=0.4482atm
In conclusion, the pressure of this gas is P=0.4482 atm.
Reference:
Chang, R. (2010). Chemistry. McGraw-Hill, New York.
