Change in enthalpy when 11.2 dm^3 of he at ntp is heated in a cylinder to 100°c is(assume ideal behaviour)
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volume of helium 11.2 dm³ or 11.2L at NTP,
we know, NTP means, gas is appeared at normal temperature and pressure.
we temperature consider 20°C
so, intimal temperature, T = 20° = 293K
given, final temperature , T' = 100°C = 373K
no of mole of gas = given volume/22.4L
= 11.2L/22.4L = 0.5
now, enthalpy change = nR∆T
where n is no of mole , R is gas constant and ∆T is change in temperature
so, enthalpy change = 0.5mol × 25/3J/mol/K × (373K-293K)
= 0.5× 25/3 × 80
= 40 × 25/3
= 1000/3
= 333.33 J
we know, NTP means, gas is appeared at normal temperature and pressure.
we temperature consider 20°C
so, intimal temperature, T = 20° = 293K
given, final temperature , T' = 100°C = 373K
no of mole of gas = given volume/22.4L
= 11.2L/22.4L = 0.5
now, enthalpy change = nR∆T
where n is no of mole , R is gas constant and ∆T is change in temperature
so, enthalpy change = 0.5mol × 25/3J/mol/K × (373K-293K)
= 0.5× 25/3 × 80
= 40 × 25/3
= 1000/3
= 333.33 J
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