0.02 , mol of an ideal diatomic gas with init | vedclass

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Question

(A) The process equation is given by $p V^2 = \beta$. Since the gas is ideal,it obeys the ideal gas equation $p V = n R T$. From the ideal gas equatio

Vedclass · Accepted Answer

$7.5 	imes 10^{-2} \, Pa \cdot m^6$

Vedclass · Answer

(A) The process equation is given by $p V^2 = \beta$. Since the gas is ideal,it obeys the ideal gas equation $p V = n R T$. From the ideal gas equation,we can write $p = \frac{n R T}{V}$. Substituting this into the process equation: $\left(\frac{n R T}{V}ight) V^2 = \beta$,which simplifies to $n R T V = \beta$. Given values: $n = 0.02 \, mol$ $R = 8.31 \, J \cdot K^{-1} \cdot mol^{-1}$ $T = 20^{\circ} C = 293 \, K$ $V = 1500 \, cm^3 = 1500 	imes 10^{-6} \, m^3 = 1.5 	imes 10^{-3} \, m^3$. Now,calculate $\beta$: $\beta = n R T V = (0.02) 	imes (8.31) 	imes (293) 	imes (1.5 	imes 10^{-3})$. $\beta = 0.02 	imes 8.31 	imes 293 	imes 0.0015 \approx 0.0729 \, Pa \cdot m^6$. Rounding to the nearest given option,$\beta \approx 7.5 	imes 10^{-2} \, Pa \cdot m^6$.

$0.02 \, mol$ of an ideal diatomic gas with initial temperature $20^{\circ} C$ is compressed from $1500 \, cm^3$ to $500 \, cm^3$. The thermodynamic process is such that $p V^2 = \beta$,where $\beta$ is a constant. Then,the value of $\beta$ is close to (the gas constant,$R = 8.31 \, J / K / mol$).

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