A sealed rigid tank contains 29 moles of an ideal gas, at an initial temperature of $270 \mathrm {~K}$ The pressure of the gas is increased until the final pressure equals 1.90 times the initial pressure. The heat capacity at constant pressure of the gas is $24.0 \mathrm {~J} / \mathrm { mol } \cdot \mathrm { K }$ How much heat is absorbed by the gas during this process? (R = 8.31 J/mol ∙ K)
A) 110 kJ
B) 170 kJ
C) 230 kJ
D) -52 kJ
E) 7.0 kJ

Question

Quizplus · Accepted Answer

Explanation:The heat absorbed by the gas is given by:$$Q = nC_p(T_f - T_i)$$where:n is the number of moles of gasCp is the heat capacity at constant pressureTi is the initial temperatureTf is the final temperatureWe are given that:n = 29 molesCp = 24.0 J/mol ∙ KTi = 270 KPf = 1.90PiWe can use the ideal gas law to find the final temperature:$$PV = nRT$$where:P is the pressureV is the volumeR is the gas constantSince the volume is constant, we have:$$\frac{Pi}{Ti} = \frac{Pf}{Tf}$$Substituting the given values, we get:$$\frac{Pi}{270 K} = \frac{1.90Pi}{Tf}$$Solving for Tf, we get:$$Tf = 270 K * 1.90 = 513 K$$Now we can calculate the heat absorbed by the gas:$$Q = nC_p(T_f - T_i)$$$$Q = 29 moles * 24.0 J/mol ∙ K * (513 K - 270 K)$$$$Q = 110 kJ$$Therefore, the correct answer is A.

A Sealed Rigid Tank Contains 29 Moles of an Ideal 270 K270 \mathrm {~K}270 K

A Sealed Rigid Tank Contains 29 Moles of an Ideal $270 \mathrm {~K}$