Question

The pressure P (in kilopascals), volume V (in liters), and temperature T (in kelvins) of a mole of an ideal gas are related by the equation P V = 8.31 T . Find the rate at which the volume is changing when the temperature is 285 K and increasing at a rate of 0.1 K/s and the pressure is 18 and increasing at a rate of 0.07 kPa/s.

Answer 1

Answer:

$\dfrac{dV}{dt} = -0.466 m^3/s$

Explanation:

given,

P (in kilo pascals),         volume V (in liters),        temperature T (in kelvins)

P V = 8.31 T

Rate of increase the temperature =  0.1 K/s

temperature = 285 K

Pressure = 18 kPa

increasing at the rate of 0.07 k Pa/s

Rate at which volume is changing = ?

$V = 8.31 \dfrac{T}{P}$

$\dfrac{dV}{dt} = 8.31 \dfrac{P\dfrac{dT}{dt}-T\dfrac{dP}{dt}}{P^2}$

$\dfrac{dV}{dt} = 8.31 \dfrac{18 \times 0.1-285\times 0.07}{18^2}$

$\dfrac{dV}{dt} = 8.31 \dfrac{-18.15}{324}$

$\dfrac{dV}{dt} = -0.466 m^3/s$