AS THE PROCESS IS ISOTHERMAL NOW AT POINT A, P= 10 * 10 ^ 5 N/M^2 , V = 10 M^3 ,
ID: 700503 • Letter: A
Question
AS THE PROCESS IS ISOTHERMAL
NOW AT POINT A, P= 10 * 10 ^ 5 N/M^2 , V = 10 M^3 , n =44KMOLE OR 44000 MOLE
USING PV = nRT PUTTING THE VALUES
(10*10^5 * 10)/(8.314 * 44000) = T
T(a) = 27.33 K
NOW SIMILARLY TEMPERATURE AT D
T(d) = PV/nR HERE P = 4*10^5 N/M^2 , V= 10 M^3
T(d) = (4*10^5 * !0)/(44000*8.314)
T(d) = 10.934 K
T(a) = T(c) BECAUSE PROCESS ISOTHERMAL SO T(c)= 27.33 K
NOW USING PV=nRT AT C POINT
P= 4* 10^5 N/M^2 , T = 27.33 , n = 44000
V= nRT/P PUTTING THE VALUES
V= (44000*8.314 * 27.33)/4*10^5
V= 24.99 M^3
SIMILARLY TEMPERATURE AT B
T(b) = PV/nR HERE P = 10*10^5 , V = 24.99 n =44000
PUTTING THE VALUES
T(b) = (10*10^5 * 24.99 )/ 44000*8.314)
T(b)= 68.313 K
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Explanation / Answer
Exercise An ideal gas undergoes the process shown in the figure. Find V2, Ta, T and Th. where n-44kmole and R-8.314J/mole.K 10x 105N/m2 4x10N/m· 1