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

Explanation:The heat absorbed by the gas is given by:$$Q = nC_v(T_f - T_i)$$where:n is the number of moles of gas$$C_v$$ is the molar heat capacity at constant volume$$T_f$$ is the final temperature$$T_i$$ is the initial temperatureWe can find the number of moles of gas using the ideal gas law:$$PV = nRT$$where:P is the pressureV is the volumeR is the ideal gas constantT is the temperatureSubstituting the given values into the ideal gas law, we get:$$(90 \mathrm { kPA })(0.60 \mathrm { m } ^ { 3 }) = n(8.31 \mathrm { J } / \mathrm { mol } \cdot \mathrm { K } )(270 \mathrm { K } )$$Solving for n, we get:$$n = 0.025 \mathrm { mol }$$Substituting the given values and the value of n into the equation for Q, we get:$$Q = (0.025 \mathrm { mol } )(28.0 \mathrm { J } / \mathrm { mol } \cdot \mathrm { K } )(270 \mathrm { K } - 460 \mathrm { K } ) = -130 \mathrm { kJ }$$Therefore, the heat absorbed by the gas is -130 kJ.

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

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