Question

A closed system consisting of 4 lb of a gas undergoes a process during which the relation between pressure and volume is pVn = constant. The process begins with p1 = 15 lbf/in.2, ν1 = 1.25 ft3/lb and ends with p2 = 60 lbf/in.2, ν2 = 0.5 ft3/lb. Determine (a) the volume, in ft3, occupied by the gas at states 1 and 2 and (b) the value of n.

Answer 1

Answer:

a) V1 = 5 ft^3

V2 = 2 ft^3

b) n = 1.378

Explanation:

Given data:

mass of gas = 4 lb

starting point

$p_1 = 15 lbf/in^2$

$v_1 = 1.25 ft^3/lb$

end point

$p_2 = 60 lbf/in^2$

$v_2 = 0.5 ft^3/lb$

Assume gas to be ideal

a) volume at point 1 $= v_1 \times m$

$V_1 = 1.25 \times 4 = 5 ft^3$

volume at point 2 $= v_2 \times m$

$V_2 = 0.5 \times 4 = 2 ft^3$

b) from ideal gas equation we have following equation

$\frac{P_1}{P_2} = [\frac{V_2}{V_1}]^n$

taking log on both side of equation

$ln [\frac{P_1}{P_2}] = n \times ln [\frac{V_2}{V_1}]$

solving for n

$n = \frac{ ln(\frac{15}{53})}{ ln(\frac{2}{5})}$

$n = \frac{-1.262}{-0.916}$

n = 1.378