Answer : The heat absorbed by the gas is closest to 34.9 kJ
Explanation :
First we have to calculate the moles of gas.
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = Pressure of gas = 70 kPa = 70000 Pa
V = Volume of gas = [tex]0.5m^3[/tex]
n = number of moles = ?
R = Gas constant = [tex]8.314m^3Pa/mol.K[/tex]
T = Temperature of gas = [tex]270K[/tex]
Putting values in above equation, we get:
[tex]70000Pa\times 0.5m^3=n\times (8.314m^3Pa/mol.K)\times 270K[/tex]
[tex]n=15.59mol[/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 = 15.59 moles
[tex]C_v[/tex] = heat capacity at constant volume gas = 28.0 J/mol.K
[tex]T_1[/tex] = initial temperature = 350 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)[/tex]
[tex]\Delta Q=(15.59moles)\times (28.0J/mol.K)\times (270-350)K[/tex]
[tex]\Delta Q=-34921.6J=-34.9kJ[/tex]
Thus, the heat absorbed by the gas is closest to 34.9 kJ
