Question

A two phase liquid vapour mixture of a pure substance has a pressure of 120 bar and occupies 0.3 $m^2$. If the mass of saturated liquid present in mixture is 2.1 kg and the mixture specific volume is 0.050 $m^2$/kg, what is the mass of vapour present in the mixture?

• A. 2.1 kg
• B. 3.1 kg
• C. 2.9 kg
• D. 3.9 kg

Answer 1

The correct option is D 3.9 kg

(d)

V= 0.3$m^3$

${}^{\nu }mixture=0.050m^3/k$

$m_f=2.1kg$
$m_g=?$
${}^{v}mixture=\frac{{}^{V}mixture}{m_f+m_g}$
$0.050=\frac{0.3}{2.1+m_g}$
$0.105+0.050m_2=0.3$
$m_2=3.9kg$