.10 mol of a monatomic gas follows the process shown below. PLEASE EXPLAIN IN DE

Question

.10 mol of a monatomic gas follows the process shown below.

PLEASE EXPLAIN IN DETAIL AND SHOW YOU WORK SO I UNDERSTAND, thank you.

Explanation / Answer

1--->2 is a isochoric that is constant volume process. So dV = 0 that is change in volume = 0 According to first law of thermodynamics, dU = dQ-dW But we know that dW=pdV But dV = 0 => dW = 0 So dU = dQ-dW = dQ-0 = dQ Cv for ideal monoatomic gas = 3R/2 So dQ = dU = nCvdT = n*3/2 R * dT = 1.5nRdT But PV = nRT So nRdT = V*(dP) and 1 atm = 1.013*(10^5) N/m^2 and 1cm^3 = 10^(-6) m^3 Substituting, dQ = 1.5*(4-2)*1.013*(10^5)*800*(10^(-6)) = 243.12J

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