Question

10 kg of air is heated in rigid vessel from 20
deg Celsius to 100 deg Celsius. If the ratio
of specific heat is 1.4 and specific gas
constant is 0.286 kJ/kgK. What will be
change in internal energy in kJ? *​

Answer 1

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Answer 2

The change in internal energy will be 800 KJ

From thermodynamics we know,

The change in internal energy (dU) of a gas is

dU = n × $C_{P}$ × dT

Where n =  number of moles = weight/ Molecular weight

$C_{P}$ = specific heat at constant pressure

dT = change in temperature

dU = (w/M) × $C_{P}$ × dT

or, dU = w × ($C_{P}$/M) × dT

or, dU = w × $C_{P}$' × dT [$C_{P}$' = molar specific heat]

Given,

w = 10 kg

dT = (100 - 20) = 80 K [As /1°C = 1 K]

γ = specific heat ratio = 1.4

Universal gas constant, R = 0.286 KJ/kgK.

Now, we know,

$C_{P}=\frac{R\gamma }{\gamma -1 }$

or,   $C_{P}$ = $\frac{0.286*1.4 }{1.4-1}$

or, $C_{P}$ = 1.001

Now, dU = 10×1×80 KJ

dU = 800 KJ

The change in internal energy will be 800 KJ.