A 2 cubic meters tank with a movable lid was initially filled with 2 moles of ga

Question

A 2 cubic meters tank with a movable lid was initially filled with 2 moles of gas at one ATM. Find the initial temperature knowing that One mole 6.023E23 particles, Boltzmann k 1.38E-23 Joule per Kelvin, One ATM-1.01ES N/sq meter The tank received a heat energy of XYZ Joules and the lid was moved to accommodate a new volume of 3 cubic meters. Find the temperature change assuming that there was no heat loss to the environment in the lid-moving process. Find the work done by the gas on the movable lid upon receiving XYZ Joules of heat. If the energy for temperature change equals to C-Temperature change and also equals to 30% of work done, calculate Cv in Joules per Kelvin. If the movable lid in the 2 cubic meters tank was fixed at all time and the initial pressure was still one ATM with the two moles of gas, find the new pressure and temperature after XYZ Joules was received by the gas with no heat loss in the pressure-rising process.

Explanation / Answer

given V = 2 m^3

movable lid

n = 2 mol

Pi = 1 atm

Ti = ?

1 mol = 6.023*10^23

k = 1.38*10^-23 J/K

1 atm = 1.01*10^5 N/m^2

hence

from ideal gas law

PiVi = n*Ao*k*Ti

Ti = 1.01*10^5*2/2*1.38*10^-23*6.023*10^23 = 12154.031287 K

b. V2 = 3 m^3

T2 = ?

Ei = XYZ joules

hence

Pi*V2 = m*Ao*k*T2

T2 = 18231.046931 K

hence

dT = (T2 - T1) = 6077.015644 K

c. work done = Pi*(dV) = 1.01*10^5*1 = 1.01*10^5 J

d. Ei = dE1 + W

dE1 = Cv*dT = 0.3*W

hence

Ei = 1.3W = 131300 J

hence

Cv = 0.3W/dT = 4.985999 J/ K

e. Ei = 131300 J

hence

Cv*dT = 131300

dT = 131300/4.98955 = 26333.734457333 K

hence

Tf = 26333.734457 + Ti = 38487.765457333 K

Pf = n*Ao*k*Tf/Vi = 3.198333*10^5 Pa = 3.1666666430 atm

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