The temperature of an ideal gas in a sealed rigid 0.60- $m ^ { 3 }$ container is reduced from 460 K to $270 \mathrm {~K}$ The final pressure of the gas is $90 \mathrm { kPA } .$ The molar heat capacity at constant volume of the gas is 28.0 J/mol ∙ K. How much heat is absorbed by the gas during this process? (R = 8.31 J/mol ∙ K)
A) -130 kJ
B) -170 kJ
C) 130 kJ
D) 170 kJ
E) 0 kJ

Question

Quizplus · Accepted Answer

Since the process occurs in a sealed rigid container, the volume is constant, and no work is done. The heat absorbed or released by the gas can be calculated using the formula $ q = nC_v\Delta T $. First, calculate the number of moles $ n $ using the ideal gas law $ PV = nRT $ with the initial conditions. Then, use the change in temperature $\Delta T = 270 \, \text{K} - 460 \, \text{K} = -190 \, \text{K}$ and the given molar heat capacity $ C_v = 28.0 \, \text{J/mol} \cdot \text{K} $ to find $ q $. The negative sign indicates heat is released, not absorbed.

The Temperature of an Ideal Gas in a Sealed Rigid m3m ^ { 3 }m3

The Temperature of an Ideal Gas in a Sealed Rigid $m ^ { 3 }$