Answer:
Thus, the heat absorbed by the gas is closest to 13.97 kJ
Explanation:
First we have to calculate the moles of gas.
Using ideal gas equation:
PV=nRT
where,
P = Pressure of gas = 70 kPa = 70000 Pa
V = Volume of gas = 0.1m^3
n = number of moles = ?
R = Gas constant = [tex]8.314m^3Pa/mol.K[/tex]
T = Temperature of gas = 270K
Putting values in above equation, we get:
[tex]70000Pa\times 0.1m^3=n\times (8.314m^3Pa/mol.K)\times 270K\\n = 3.118mol[/tex]
Heat released at constant volume is known as internal energy.
The formula used for change in internal energy of the gas is:
[tex]:\Delta Q=\Delta U\\\\\Delta U=nC_v\Delta T\\\\\Delta Q=nC_v(T_2-T_1)[/tex]
where,
[tex]\Delta Q[/tex] = heat at constant volume = ?
[tex]\Delta U[/tex] = change in internal energy
n = number of moles of gas = 3.118 moles
[tex]C_v[/tex]= heat capacity at constant volume gas = 28.0 J/mol.K
[tex]T_1[/tex]= initial temperature = 430 K
[tex]T_2[/tex] = final temperature = 270 K
Now put all the given values in the above formula, we get:
[tex]\Delta Q=nC_v(T_2-T_1)\Delta Q=(3.118moles)\times (28.0J/mol.K)\times (270-430)K\Delta Q=-13970.19J\\\\=-13.97kJ\\[/tex]
Thus, the heat absorbed by the gas is closest to 13.97 kJ
