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

IF U SATISFY WITH THE SOLUTION PLEASE LIKE IT IT ..............THNKXXXXXX

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

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