The pressure p and the volume v of a gas are connected by the relation pv1.4 = const. Find the percentage error in p corresponding to a decrease of
in v.
Given pv1.4 = constant and the decrease in v is
.
Hence, we have
∴ Δv = –0.005v
We have pv1.4 = constant
Taking log on both sides, we get
log(pv1.4) = log(constant)
⇒ log p + log v1.4 = 0 [∵ log(ab) = log a + log b]
⇒ log p + 1.4 log v = 0 [∵ log(am) = m log a]
On differentiating both sides with respect to v, we get
We know
Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as
Here, 
and Δv = –0.005v
⇒ Δp = (–1.4p)(–0.005)
∴ Δp = 0.007p
The percentage error is,
⇒ Error = 0.007 × 100%
∴ Error = 0.7%
Thus, the error in p corresponding to the decrease in v is 0.7%.