If a closed container of volume 200 lit. of O_{2} gas (ideal gas) at 1

1.	If a closed container of volume 200 lit. of O_{2} gas (ideal gas) at 1 atm & 200 K is taken to planet. Find the pressure of oxygen gas at the planet at 821 K in same container(A)10/(e ^ 100)(B)20/(e ^ 50)(C) 1 atm (D) 2 atmhansion #1 (0 No. 8 to 10)
Answer» We have given, Volume of container = 200 litre initial pressure of g oxygen gas = 1 atm initial temperature = 200 K Temperature of planet (final Temperature) = 821 K ∵ Oxygen gas in the same container, it means. Volume of gas constant. According to Gay-Lussac law — At constant volume, As pressure increases, temperature also increases. P ∝ T or P₁/T₁ = P₂/T₂ where, P₁ = initial pressure T₁ = initial Temperature P₂ = final pressure T₂ = final Temperature Therefore, \(\frac{1\,atm}{200\,K}=\frac{P_2}{821\,K}\) P₂ = \(\frac{821\,K\times1\,atm}{200\,K}\) P₂ = 4.105 atm.

If a closed container of volume 200 lit. of O_{2} gas (ideal gas) at 1 atm & 200 K is taken to planet. Find the pressure of oxygen gas at the planet at 821 K in same container(A)10/(e ^ 100)(B)20/(e ^ 50)(C) 1 atm (D) 2 atmhansion #1 (0 No. 8 to 10)

Answer»

We have given,

Volume of container = 200 litre

initial pressure of g oxygen gas = 1 atm

initial temperature = 200 K

Temperature of planet (final Temperature) = 821 K

∵ Oxygen gas in the same container, it means.

Volume of gas constant.

According to Gay-Lussac law — At constant volume,

As pressure increases, temperature also increases.

P ∝ T

or P₁/T₁ = P₂/T₂

where,

P₁ = initial pressure

T₁ = initial Temperature

P₂ = final pressure

T₂ = final Temperature

Therefore,

\(\frac{1\,atm}{200\,K}=\frac{P_2}{821\,K}\)

P₂ = \(\frac{821\,K\times1\,atm}{200\,K}\)

P₂ = 4.105 atm.

Discussion

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