In a cylinder,1.0 , text{kcal} of heat is add | vedclass

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Question

(A) Given: $q = +1.0 \, 	ext{kcal} = 1000 \, 	ext{cal}$ (heat added to the system).
Pressure $P = 1 \, 	ext{atm}$.
Change in volume $\Delta V = V

Vedclass · Accepted Answer

$0.99274$

Vedclass · Answer

(A) Given: $q = +1.0 \, 	ext{kcal} = 1000 \, 	ext{cal}$ (heat added to the system).
Pressure $P = 1 \, 	ext{atm}$.
Change in volume $\Delta V = V_f - V_i = 1.5 \, 	ext{L} - 1.2 \, 	ext{L} = 0.3 \, 	ext{L}$.
Work done $W = -P \Delta V = -1 \, 	ext{atm} 	imes 0.3 \, 	ext{L} = -0.3 \, 	ext{L} \cdot 	ext{atm}$.
Conversion factor: $1 \, 	ext{L} \cdot 	ext{atm} = 24.22 \, 	ext{cal}$.
So,$W = -0.3 	imes 24.22 \, 	ext{cal} = -7.266 \, 	ext{cal} = -0.007266 \, 	ext{kcal}$.
Using the first law of thermodynamics: $\Delta U = q + W$.
$\Delta U = 1.0 \, 	ext{kcal} - 0.007266 \, 	ext{kcal} = 0.992734 \, 	ext{kcal} \approx 0.99274 \, 	ext{kcal}$.

In a cylinder,$1.0 \, \text{kcal}$ of heat is added to $1.2 \, \text{L}$ of $O_2$ gas at a constant pressure of $1 \, \text{atm}$. The volume changes to $1.5 \, \text{L}$. Calculate the change in internal energy $(\Delta U)$ for this process. (in $, \text{kcal}$)

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