Gas equation. Pressure in a gastank 10bars, from 20C to 90C

Question

The pressure in a gastank is 10bars. Calculate the pressure in the gastank as the temprature goes from 20C to 95C

 

My calculations:

273.15+95= 368.15K

273.15+20= 293.15K

 

Convert 10bars into 1atm.

 

Then use p1 * v1 / t1 = p2 * v2 /t2

 

1atm * v / 293.15 = P2 * v2 / 368k .... Becomes: 1atm /293.15k = P2 /368K? Is this right?

Answer 1

There is nothing in the problem which says that the answer should be given in units of atm, so the usual assumption should be that the pressure units  should be kept in the units given for the starting pressure.  The volume of the tank is fixed, so it doesn't need to be considered in the calculations.  According to Amonton's Law (aka Gay-Lussac's Law), the pressure and Kelvin temperature of an ideal gas at constant volume are directly proportional.  Change in pressure and/or Kelvin temperature are related by the equation P1/T1 = P2/T2.  In terms of P2, P2 = (P1/T1)(T2).  For this problem, P2 = (10 bar/293.15 K)(368.15) = 12.6 bar.  If you had to convert to atm, just multiply by 1.01325 ( = 12.7 atm).  Technically, if the starting pressure is known to only one significant figure (and presumably, the temperatures are known to the nearest degree), the final answer should only have one significant figure (i.e., be 10 bar)