A mass of diatomic gas (gamma = 1.4) at a pre | vedclass

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(D) For an adiabatic process,the relationship between temperature and pressure is given by:$\frac{T^\gamma}{P^{\gamma-1}} = 	ext{constant}$Therefore,

Vedclass · Accepted Answer

$256$

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(D) For an adiabatic process,the relationship between temperature and pressure is given by:$\frac{T^\gamma}{P^{\gamma-1}} = 	ext{constant}$Therefore,$\left(\frac{T_i}{T_f}ight)^\gamma = \left(\frac{P_i}{P_f}ight)^{\gamma-1}$,which implies $P_f = P_i \left(\frac{T_f}{T_i}ight)^{\frac{\gamma}{\gamma-1}}$  ...$(i)$Given:$T_i = 27^{\circ}C = 300 	ext{ K}$$T_f = 927^{\circ}C = 1200 	ext{ K}$$P_i = 2 	ext{ atm}$$\gamma = 1.4$Substituting these values into equation $(i)$:$P_f = 2 	imes \left(\frac{1200}{300}ight)^{\frac{1.4}{1.4-1}}$$P_f = 2 	imes (4)^{\frac{1.4}{0.4}}$$P_f = 2 	imes (4)^{3.5}$$P_f = 2 	imes (2^2)^{3.5} = 2 	imes 2^7 = 2^8 = 256 	ext{ atm}$.

$A$ mass of diatomic gas $(\gamma = 1.4)$ at a pressure of $2 \text{ atm}$ is compressed adiabatically so that its temperature rises from $27^{\circ}C$ to $927^{\circ}C$. The pressure of the gas in the final state is ...... $\text{atm}$.

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