A closed copper vessel contains water equal to half of its volume. When the temperature of the vessel is raised to 447∘C, the pressure of the steam in the vessel is (treat stream as an ideal gas, R=8310 JK–1/kg mole, density of water =1000 kgm–3 molecular weight of water = 18)

PV = nRTPV=mMRT⇒PV=V2ρMRT⇒P=10002×18×8310×720⇒P=16.62×107Pa

A closed copper vessel contains water equal to half of its volume. When the temperature of the vessel is raised to 447∘C, the pressure of the steam in the vessel is (treat stream as an ideal gas, R=8310 JK–1/kg mole, density of water =1000 kgm–3 molecular weight of water = 18)

PV = nRTPV=mMRT⇒PV=V2ρMRT⇒P=10002×18×8310×720⇒P=16.62×107Pa

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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 the steam in the vessel is (treat stream as an ideal gas, $R = 8310 J K^{- 1} / k g$ mole, density of water $= 1000 k g m^{- 3}$ molecular weight of water = 18)

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$33.24 \times 10^{7} P a$

$8.31 \times 10^{7} P a$

$10.31 \times 10^{7} P a$

$16.62 \times 10^{7} P a$

answer is B.

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Detailed Solution

PV = nRT

$PV = \frac{m}{M} RT \Rightarrow PV = \frac{\frac{V}{2} ρ}{M} RT$

$\Rightarrow P = \frac{1000}{2 \times 18} \times 8310 \times 720$

$\Rightarrow P = 16.62 \times 10^{7} Pa$

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A closed copper vessel contains water equal to half of its volume. When the temperature of the vessel is raised to 447∘C, the pressure of the steam in the vessel is (treat stream as an ideal gas, R=8310 JK–1/kg mole, density of water =1000 kgm–3 molecular weight of water = 18)

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