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−1​​R​​(T​i​​−T​f​​)

Here

R = Universal gas constant

γ= Ratio of molar specific heat

T​i​​= Initial absolute temperature

T​f​​= Final absolute temperature

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• Now the equation for a adiabatic compression is PV​γ​​ = constant

Now we know PV=nRT

=> P=nRT/V

Substituting the value of P, we have

​V​​nRT​​∗V​γ​​=constant

=> nTV​γ−1​​=constant

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Here '1' represent initial condition & '2' represent final condition of the air.

So we have

nT​1​​V​1​γ−1​​=nT​2​​V​2​γ−1​​

=> T​2​​= T​1​​∗ (​V​2​​​​V​1​​​​)​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, T​2​​= 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−1​​R​​(T​1​​−T​2​​) = 0.3 mole x ​1.4−1​​8.31​​ ​mole. K​​J​​ 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.