The temperature of an ideal gas in a sealed rigid 0.60-
Container is reduced from 460 K to
The final pressure of the gas is
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

The temperature of an ideal gas in a sealed rigid 0.60- 
Container is reduced from 460 K to 
The final pressure of the gas is 
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

Quizplus · Accepted Answer

We can use the formula for the change in internal energy of an ideal gas:
ΔU = (3/2) nRΔT
where n is the number of moles of gas and ΔT is the change in temperature. Since the container is rigid, there is no work done by the gas, so all the energy exchanged is in the form of heat. Therefore,
Q = ΔU = (3/2) nRΔT
To find n, we can use the ideal gas law:
PV = nRT
n = PV/RT
Substituting this expression for n into the equation for Q, we get:
Q = (3/2) PVΔT/R
Plugging in the given values, we get:
Q = (3/2) (11.0 atm)(0.60 L)(1.01325 × 10^5 Pa/atm)(460 K - 110 K)/(8.31 J/mol ∙ K)
Q = -130 kJ (rounded to three significant figures) 
The negative sign indicates that heat is absorbed by the gas. Therefore, the correct choice is A) -130 kJ.

The Temperature of an Ideal Gas in a Sealed Rigid