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A vessel of volume V = 6.0 1 contains water together with its saturated vapour under a pressure of 40 atm and at a temperature of 250 ^@ C. The specific volume of the vapour is equal to V'_v = 50 1//kg under these conditions. The total mass of the system water-vapour equals m = 5.0 kg. Find the mass and the volume of the vapour. |
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Answer» Solution :The specific volume of water (the liquid) will be WRITTEN as `V'_l` since `V'_v GT gt V'_l` most of the weight is due to water. Thus if `m_l` is mass of the liquid and `m_v` that of the VAPOUR then `m = m_l + m_v` `V = m_l V'_l + m_v V'_v` or `V - m V'_l = m_v(V'_v = V'_l)` So `m_v =(V - mV'_l)/(V'_v - V'_1) = 20 gm` in the present case. Its volume is `m_v V'_v = 1.01`. |
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