When 80 J of heat is absorbed by a monoatomi | vedclass

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(A) For a monoatomic gas,the degrees of freedom $f = 3$. At constant pressure,the heat absorbed is given by $Q = n C_p \Delta T$. Since $C_p = \frac{f

Vedclass · Accepted Answer

$2 	imes 10^5 \ Nm^{-2}$

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(A) For a monoatomic gas,the degrees of freedom $f = 3$. At constant pressure,the heat absorbed is given by $Q = n C_p \Delta T$. Since $C_p = \frac{f+2}{2} R = \frac{5}{2} R$,we have $Q = \frac{5}{2} n R \Delta T$. From the ideal gas equation,$p \Delta V = n R \Delta T$. Substituting this into the heat equation: $Q = \frac{5}{2} p \Delta V$. Given $Q = 80 \ J$ and $\Delta V = 16 	imes 10^{-5} \ m^3$,we have: $80 = \frac{5}{2} 	imes p 	imes 16 	imes 10^{-5}$. $80 = 40 	imes 10^{-5} 	imes p$. $p = \frac{80}{40 	imes 10^{-5}} = 2 	imes 10^5 \ Nm^{-2}$.

When $80 \ J$ of heat is absorbed by a monoatomic gas,its volume increases by $16 \times 10^{-5} \ m^3$ at constant pressure. The pressure of the gas is

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