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Chemistry

What temperature would be necessary to double the volume of a gas initially at s.t.p. if the pressure is decreased by 50% ?

Gas Laws

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Answer

Initial conditions [S.t.p.]Final conditions
P1 = Initial pressure of the gas = 1 atmP2 = Final pressure of the gas = 0.5 atm
V1 = Initial volume of the gas = VV2 = Final volume of the gas = 2V
T1 = Initial temperature of the gas = 273 KT2 = Final temperature of the gas = ?

\ By Gas Law:

P1×V1T1=P2×V2T2\dfrac{\text{P}1\times\text{V}1}{\text{T}1} = \dfrac{\text{P}2\times\text{V}2}{\text{T}2}

Substituting the values :

1×V273=0.5×2VT2T2=0.5×2×273T2=273K\dfrac{1\times \text{V}}{273} = \dfrac{0.5\times\text{2\text{V}}}{\text{T}2} \\[0.5em] \text{T}2 = 0.5\times 2 \times 273\\[0.5em] \text{T}_2 = 273\text{K}

Therefore, final temperature of the gas = 273 K = 0 °C

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