The value of q, w, ΔU and ΔH are +2232J, -2232J, 0, 0
Data given;
Using ideal gas equation, we can calculate the volume
PV = nRT
substituting the values
[tex]PV=nRT\\\\10*v = 1*8.314*10^-^5*298\\V_1= 2.48*10^-^3m^3[/tex]
The initial volume is 2.48*10^-3 m^3
However, the volume increased by 10 times.
[tex]v_2= 2.48*10^-^3*10=0.0248m^3[/tex]
The work done(w) on the gas is
[tex]W=-p\delta V\\W= -1*(0.0248-0.00248)\\W=-0.02232barm^3\\W=-2232J[/tex]
NB; 1 bar m^3 = 1000J
[tex]\delta U= C_v\delta T\\\\\delta U = 0[/tex]
This is an isothermal process, i.e a process that underwent no change in temperature.
[tex]\delta U = Q+W\\0=Q+W\\Q= -W\\Q = -(-2232)\\Q = 2232J[/tex]
The value of Q is 2232J
The ΔH of the gas can be calculated as
[tex]\delta H = C_p\delta T = 0[/tex]
Since there's no change in temperature, ΔH = 0
From the calculations above, the value of q, w, ΔU and ΔH are +2232J, -2232J, 0, 0
Learn more on work done on a gas and isothermal process here;
https://brainly.com/question/16014998
