Answer:
5118.50 J
Step-by-step explanation:
pΔv=nRΔT ;
therefore, ΔT=PV/nR
ΔT = (6.4×10^5)(3.2×10^(-3)/1×8.314
ΔT= 2.4633×10^2 = 246.33 K
specific heat at constant pressure is given as:
c_p = 3/2R
c_p = 12.5 J/mol K
Now, substitute in equation (1)
we know that
Q=ΔU+W ;
and
W=pΔV= 6.4×10^5×3.2×10^(-3) = 2048 J
now
ΔU=CvΔT = 12.465×246.33 =3070.50 J ;
therefore
Q=3070.50+2048= 5118.50 J