physics
posted by Anonymous .
One kilogram of an ideal gas is heated from 27*C to 327*C. If the volume of the gas remains constant, the ratio of the pressure of the gas before heating to the pressure after is
a. 1:4
b. 1:3
c. 1:2
d. 1:1
Work:
1atm/293K=P/(327+273)
P=2 So, from what I got, I think the answer is c.

Yes, the absolute temperature T doubles. Since V is constant, the pressure also doubles. The answer is (c)
Your formula should be
P1/T1 = {P2/T2)
P2/P1 = T2/T1 = (327+273)/(27+273)
= 600/300
P1 does not have to be 1 atm, and you did not calculate T1 correctly. It is 300, not 293.