Q7 During one compression stroke ( process) in an engine , 0.30 moles of air ( a diatomic ideal gas ) are compressed adiabatically from an initial volume and temperature of V1= 5.3 L and T1=277K to fi
Answer 7
The magnitude of work done by the gas during this process = 1651.806
Explanation:
We know, For a adiabatic compression, Work done ( For 1 mole gas ) = γ−1R(Ti−Tf)
Here
R = Universal gas constant
γ= Ratio of molar specific heat
Ti= Initial absolute temperature
Tf= Final absolute temperature
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Now the equation for a adiabatic compression is PVγ = constant
Now we know PV=nRT
=> P=nRT/V
Substituting the value of P, we have
VnRT∗Vγ=constant
=> nTVγ−1=constant
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Here '1' represent initial condition & '2' represent final condition of the air.
So we have
nT1V1γ−1=nT2V2γ−1
=> T2= T1∗ (V2V1)γ−1
Here
V1= 5.3 L
V2=0.9 L and T1=277 K
γ= 7/5 = 1.4 ( We know for diatomic gas this ratio is 1.4 )
=> Final temperature, T2= 277 K x (5.3/0.9)1.4 = 11.969 K
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Work done for n mole of air = n∗ γ−1R(T1−T2) = 0.3 mole x 1.4−18.31 mole. KJ x ( 277 - 11.969 ) K = 1651.806 Joules
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Correction : Final temperature, T_2 = 277 K x (5.3/0.9)^1.4 = 3315.33 K
Work done to compress the air = n ∗ R/γ−1 * (T1 −T2 ) = 0.3 x 8.31 /1.4−1 x ( 277 - 3315.33 ) = -18936.39173 Joules
More accurate answer is 1782.3715 ( Rounded up to 4 decimal place). This is final answer.