Gaseous hydrogen contained initially under standard condition in a sealed vessel of volume 5 litre was cooled by - T -55 K- Find by how much the internal energy of the gas in joules will change- Answer the nearest multiple of 50

Because of sealed vessel nbsp;Δ V = 0 So, Δ W = 0 from ideal gas equation nbsp; Δ Q = Δ U + Δ W nbsp; Δ Q = Δ U we know that Δ U = nC_v Δ T = mM×f2 R ...

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Standard XII

Chemistry

Mean Free Path, Collision Frequency

Gaseous hydro...

Question

Gaseous hydrogen contained initially under standard condition in a sealed vessel of volume 5 litre was cooled by $Δ T = 55$ K. Find by how much the internal energy of the gas in joules will change. Answer the nearest multiple of 50

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Solution

Because of sealed vessel $Δ V = 0$
So, $Δ W = 0$
from ideal gas equation -
$Δ Q = Δ U + Δ W$
$Δ Q = Δ U$
we know that -
$Δ U = n C_{v} Δ T = \frac{m}{M} \times \frac{f}{2} R Δ T$
$= \frac{m}{M} \times \frac{5}{2} R Δ T . . . . . (1)$
ideal gas equation $- PV = nRT = \frac{m}{M} R T$
$= \frac{m}{M} R = \frac{P V}{T}$
put the value of $\frac{m}{M}$ in equation -
$Δ U = Δ Q = \frac{P V}{T} \times \frac{5}{2} Δ T$
$= \frac{(1.013 \times 10^{5}) (5 \times 10^{- 3})}{273} \times \frac{5}{2} \times (- 55)$
=-255 J

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Gaseous hydrogen contained initially under standard condition in a sealed vessel of volume 5 litre was cooled by ΔT=55 K. Find by how much the internal energy of the gas in joules will change. Answer the nearest multiple of 50

Because of sealed vessel ΔV=0 So, ΔW=0 from ideal gas equation - ΔQ=ΔU+ΔW ΔQ=ΔU we know that - ΔU=nCvΔT=mM×f2RΔT =mM×52RΔT.....(1) ideal gas equation - PV = nRT=mMRT =mMR=PVT put the value of mM in equation - ΔU=ΔQ=PVT×52ΔT =(1.013×105)(5×10−3)273×52×(−55) =-255 J

Gaseous hydrogen contained initially under standard condition in a sealed vessel of volume 5 litre was cooled by $Δ T = 55$ K. Find by how much the internal energy of the gas in joules will change. Answer the nearest multiple of 50