a.what is the final temperature?

Question

Helium gas with a volume of 2.30 L, under a pressure of 1.00 atm and at a temperature of 41.0°C, is warmed until both pressure and volume are doubled.

Answer 1

As Boyle's Law indicates, the product P V is proportional to T. So, if BOTH P and V are doubled, the ABSOLUTE temperature is multiplied by 4.

This implies it's finally 1256.6 K or 983.45 C (***: but see foootnote)

Here's the calculation:

0 deg C = 273.15 K, so 41.0 C = 314.15 K (***: but see footnote)

4 x 314.15 K = 1256.6 K or 983.45 C QED

Live long and prosper.


*** Actually there is a small uncertainty here, since you didn't specify whether "C" meant "Centigrade" or "Celsius." It's unfortunate that the symbol "C" is used for both --- their "degrees" are the same size, but their zero points DIFFER, if only by 0.01 C. Absolute zero K is at
-273.15 CENTIGRADE, but at -273.16 CELSIUS. The Celsius scale is based on 0° at the triple point of water (0.01°Centigrade) and centigrade has 0° at the freezing point of water.

I answered you as though you had meant "Centigrade." If you meant "Celsius," then the final temperatures would have been 1256.64 K or 983.48 Celsius. hardly a very significant difference, but slightly different nonetheless,

Answer 2

p1v1/n1r1t1 = p2v2/n2r2t2

n and are are constant throughout the process, so they can drop out

p1v1/t1 = p2v2/t2

p1v1/t1 = 2.3/314.15 (K) convert 41 celsius to kevlin

2.3/314.15 = p2v2/t2

2.3/314.15 = .007321

.007321 = p2v2/t2

p2v2 = 4.6 * 2

.007321 = 9.2/t2

t2 = 9.2/.007321

t2 = 1256.66 K