Question

A 2.60-mol sample of helium gas initially at 300 K, and 0.400 atm is compressed isothermally to 1.00 atm. Note that the helium behaves as an ideal gas.

(a) Find the final volume of the gas.


(b) Find the work done on the gas.

(c) Find the energy transferred by heat.

Answer 1

Given number of mol., n = 2.6 mol Initial temperature, Ti = 300k = T Initial pressure, P = 0.4 atm. Final pressure, P = 1 atm value of Gas Constant R = 0.082057 Latm.mol (a) For ideal gas, K7 Pf V = nRT; V - MRT =Y 2 v . 2:6 x 0:082057 x 300, So, the is final volume of the gas 64 liter. on gas in isothermal work process done is W-VRT 2.6 x 8.314 x 300 m (0.4) 5 I w = – 5.942 x 10 )

So, gas work done on the is 5.942 X10 J In isothermal : process, a temperature remain Constant So internal energy Change, 00 = 0 So, from 1st law thermodynamic, Q = AU + W => Q = 0 = -5.942 KJ So, energy transferred by heat is Q = 5.942 XJ.
Thank you.