During an adiabatic compression,830 J of wor | vedclass

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(C) Given: Work done on the gas,$W = -830 \ J$ (since work is done on the system).Number of moles,$\mu = 2$.For a diatomic ideal gas,the adiabatic exp

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$20$

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(C) Given: Work done on the gas,$W = -830 \ J$ (since work is done on the system).Number of moles,$\mu = 2$.For a diatomic ideal gas,the adiabatic exponent $\gamma = 1.4$.Work done in an adiabatic process is given by the formula:$W = \frac{\mu R (T_1 - T_2)}{\gamma - 1} = -\frac{\mu R \Delta T}{\gamma - 1}$Substituting the values:$-830 = -\frac{2 	imes 8.3 	imes \Delta T}{1.4 - 1}$$830 = \frac{16.6 	imes \Delta T}{0.4}$$830 = 41.5 	imes \Delta T$$\Delta T = \frac{830}{41.5} = 20 \ K$Thus,the change in temperature is $20 \ K$.

During an adiabatic compression,$830 \ J$ of work is done on $2 \ moles$ of a diatomic ideal gas to reduce its volume by $50\%$. The change in its temperature is nearly..... $K$ $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$

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