An ideal monatomic gas is contained in a vessel of constant volume 0.120 m3. The
ID: 1622802 • Letter: A
Question
An ideal monatomic gas is contained in a vessel of constant volume 0.120 m3. The initial temperature and pressure of the gas are 300 K and 5.00 atm, respectively. The goal of this problem is to find the temperature and pressure of the gas after 26.0 kJ of thermal energy is supplied to the gas. (a) Use the ideal gas law and initial conditions to calculate the number of moles of gas in the vessel. Correct: Your answer is correct. . mol (b) Find the specific heat of the gas. Incorrect: Your answer is incorrect. . Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. J/K (c) What is the work done by the gas during this process? Correct: Your answer is correct. . kJ (d) Use the first law of thermodynamics to find the change in internal energy of the gas. Correct: Your answer is correct. . kJ (e) Find the change in temperature of the gas. Correct: Your answer is correct. . K (f) Calculate the final temperature of the gas. K (g) Use the ideal gas expression to find the final pressure of the gas. atm
Explanation / Answer
Volume of the gas (V) = 0.120 m³
Temperature of the gas (T1) = 300 K
Pressure of the gas (P1) = 5 atm
Thermal energy supplied at constant volume (nCvdt) = 26 kJ
(a) from ideal gas equation PV = nRT
5 × 1 × 105 × 0.120 = n × 8.314 × 300
n = 24 mol
(b) Specific heat at constant volume (Cv) = R/(-1)
for monoatomic = 1.6
So Cv = 8.314 / (1.6 -1)
Cv = 13.86 J/molK
(specific heat * molar mass = molar specific heat)
(c) since volume is constant so the work done during the process = PV
Work done = 0
(d) from the first law of thermodynamics
Q = U + W
26000 = U + 0
So U = 26000 J or 26 kJ
(e) heat supplied (Q) = nCvdT
26000 = 24 × 13.86 × dt
dt = 78 K
(f) from ideal gas equation
P1V1/T1 = P2V2/T2
5 / (300) = P2/ (300+78)
P2 = 6.3 atm
#since molar mass is not given in the question so we can't find specific heat.
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