Question

Use the ideal gas law Pwith volume V in cubic inches (in), temperature 7' in Kelvins (K) and R = 10(inIb / K) to find the rate at which the temperature of a gas is changing when the volume is 200in" and increasing at the rate of 4in/s, while the pressure is 5lb/in² and de creasing at the rate of 11b/in/s.

Answer 1

Solution;

Given;

R=10(inlb/K)

V=200in³

$\frac{dV}{dt}=4in/s$

P=5lb/in²

$\frac{dP}{dt}=-11lb/in/s$

The ideal gas law is;

$PV=nRT$

Differentiate both sides with respect to t;

$\frac{d(PV)}{dt}=\frac{d(nRT)}{dt}$

$V\frac{dP}{dt}+P\frac{dV}{dt}=nR\frac{dT}{dt}$

Hence;

$\frac{dT}{dt}=\frac{V\frac{dP}{dt}+P\frac{dV}{dt}}{nR}$

Take n=1,by substitution;

$\frac{dT}{dt}=\frac{(200×-11)+(5×4)}{1×10}$

$\frac{dT}{dt}=-218K/s$

Hence;

The temperature is decreasing at the rate of $218K/s$ .