Question

The temperature $T$ and the volume $V$ of a gas are connected by the relation $T{V}^{1.2}=$ constant. The percentage error in $T$ which corresponds to a decrease of $\frac{1}{5}\%$ in $V$ is

• A. $0.20$
• B. $0.24$
• C. $0.70$
• D. $0.76$

Answer 1

The correct option is B $0.24$

$\frac{dV}{V}\times 100=-\frac{1}{5}$
Also, ${TV}^{1.2}=k$ (say)
Applying $\ln$ on both sides, we get
$\ln T+1.2\ln V=\ln k$
Differentiating both sides, we get
$\frac{dT}{T}+1.2\frac{dV}{V}=0$
$\Rightarrow \frac{dT}{T}=-1.2\frac{dV}{V}$
$\Rightarrow \frac{dT}{T}\times 100=-1.2\times \frac{dV}{V}\times 100$
$\Rightarrow \frac{dT}{T}\times 100=-1.2\times (-\frac{1}{5})=0.24$
$\therefore$ Percentage error in $T=0.24$