Question

Three moles of an ideal gas are inside a 5.0 L chamber. 50.6 kJ of heat are added to the gas and, in the process, the pressure increases from 2.0 atm to 10.0 atm.

(A) Find the initial and final temperatures of the gas (in both °C and K)

(B) Find the change in internal energy of the gas

Answer 1

since volume is constant = 5L = 5*10^-3 m³

1 atm =101325 pascals

PV= nRT

T_i = PV/nR = 2* 101325*5*10^-3 / ( 3*8.314)

=40.6242483 K  or 40.6242483 -273 = -232.375752 deg celsius

Volume is constant so  

P_i/T_i = P_f /T_f

2/ 40.6242483 = 10 / T_f

T_f = 10 / (2/ 40.6242483 ) = 203.121241 K or  203.121241 -273 = -69.878759 deg celsius

(B)

Q = ΔU+W ( volume is constant so W= 0

ΔU = Q = 50.6 kJ answer

let me know in a comment if there is any problem or doubts​