the gas is enclosed in a cylinder under STP conditions at what temperature does the volume of enclosed gas become one sixth of its initial volume pressure remaining constant
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In present case , pressure is constant. Thus ideal gas equation can be rewritten as
v/T = mR/p = constant ( since 'mass' and 'characteristic gas constant' are going to be remain same)
Thus,
T2/T1 = v2/v1 ( but in question it was given, v2 = v1/6 & T1=20°C i.e STP)
That is,T2/(20+273) = v1/6×v1 ( 273 is added with 20°C to convert in Kelvin).
T2=293/6= 48.33 K i.e. -224°C( that is your final temperature)
v/T = mR/p = constant ( since 'mass' and 'characteristic gas constant' are going to be remain same)
Thus,
T2/T1 = v2/v1 ( but in question it was given, v2 = v1/6 & T1=20°C i.e STP)
That is,T2/(20+273) = v1/6×v1 ( 273 is added with 20°C to convert in Kelvin).
T2=293/6= 48.33 K i.e. -224°C( that is your final temperature)
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