a cylinder of fixed capacity 67.2 litres contains helium gas at st.p.the amount of heat required to raise the temperature of the gas by 15 degree celsius is(R=8.31 j/mol/k) a.520j b.560.9j c.620j d.621.2j plz explain with solution
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1
Volume V = 67.2 litres of Helium at STP
ΔT = 15⁰K
1 mole of an ideal gas occupies 22.4 litres at STP.
So number of moles of Helium gas n = 67.2/22.4 = 3
we know that
γ = ratio of specific heats for an ideal mono atomic gas = 5/3
C_p = heat capacity at constant pressure for a mono atomic gas = 5/2 R
C_v = heat capacity at constant volume for a mono atomic gas = 3/2 R
Change in internal energy = n C_v ΔT
= 3 * 3/2 * 8.31 J/mol/°K * 15°K
= 560.925 Joules
There is no change in the volume and hence no work done in heating the gas by 15 deg.
Heat supplied = change in internal energy
= 560.925 Joules
ΔT = 15⁰K
1 mole of an ideal gas occupies 22.4 litres at STP.
So number of moles of Helium gas n = 67.2/22.4 = 3
we know that
γ = ratio of specific heats for an ideal mono atomic gas = 5/3
C_p = heat capacity at constant pressure for a mono atomic gas = 5/2 R
C_v = heat capacity at constant volume for a mono atomic gas = 3/2 R
Change in internal energy = n C_v ΔT
= 3 * 3/2 * 8.31 J/mol/°K * 15°K
= 560.925 Joules
There is no change in the volume and hence no work done in heating the gas by 15 deg.
Heat supplied = change in internal energy
= 560.925 Joules
Answered by
1
Answer:
Explanation:
q=ncvdt
67.2\22.4X3R\2X20=90R
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