The pressure of one mole of an ideal gas of volume 10\(^-2\) m\(^3\) at a temperature of 27°c is
[molar gas constant = 8.3 J mol\(^{-1}\) k\(^{-1}\)]
A
2.49 x 104 Nm-2
B
2.24 x 104 Nm-2
C
2.49 x 105 Nm-2
D
2.24 x 105 Nm-1
Verified Correct Answer: C.
Physical Principles & Numerical Solution
T = 27 + 273 = 300 K, volume, V = 10-2 m3 molar gass constant, R = 8.3 Jmol- 1k - 1 number of mole, n = 1
PV = NRT
i.e P = (NRT)/V = (1 x 8.3 x 300)/10-2
= 2.49 x 10\(^5\) Nm\(^-2\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #14313)