Question

A closed copper vessel contains water equal to half of its volume when the temperature. Of the vessel is raised to ${447^ \circ }C$ the pressure of steam in the vessel is ( Treat steam as an ideal gas, $R = 8.310J{K^{ - 1}}mo{l^{ - 1}}$ , density of water =$1000kg{m^3}$ , molecular weight of water =$18$ )
A. $33.24 \times {10^7}Pa$
B. $16.62 \times {10^7}Pa$
C. $10.31 \times {10^7}Pa$
D. $8.31 \times {10^7}Pa$

Answer 1

Hint We will use the ideal gas equation to serve the purpose.
i.e. $PV = nRT$
where, P= pressure of steam in the vessel.
V= volume of vessel.
R= gas constant.
T= temperature of steam in the vessel.
n= number of moles.
Further we will use basic expression of density i.e. mass = density x volume to calculate the mass of gas filled in the vessel.

Complete Step By Step Solution If V be the volume of vessel and steam is treated as ideal gas then from ideal gas equation we have,
$PV = nRT$
Now we know, $n = \dfrac{m}{M}$
Therefore, $PV = \dfrac{m}{M}RT......(1)$
Where, m is given mass and M is molecular mass.
Now, mass of liquid in vessel =density of liquid x volume
i.e. $m = \rho \times \dfrac{V}{2}......(2)$
Now using equation (1) and (2) we get,
$PV = \dfrac{{\rho V}}{{2M}}RT$
Eliminating V from both sides, we get
$P = \dfrac{{\rho RT}}{{2M}}$
Now putting values in above equation, we get
$P = \dfrac{{8.310 \times 447 \times 1000}}{{2 \times 18}}$
On solving we get
$P = 16.62 \times {10^7}Pa$
Hence the required pressure is $P = 16.62 \times {10^7}Pa$

Option (B) is correct.

Note $PV = nRT$ is valid only for ideal gases. Here we have considered steam as ideal gas. If attractive forces come into play then this expression won’t work. We have to use expression for real gases i.e. $\left( {P + \dfrac{{a{n^2}}}{{{V^2}}}} \right)(v - b) = nRT$