Answer
(a) $1079.66J$
(b) $24.06K$
(c) $6.00\times 10^{-3}m^3$
Work Step by Step
(a) We can find the internal energy as follows:
$\Delta T=\frac{2Q}{5nR}$
$\Delta T=\frac{2(1800J)}{5(3.6mol)(8.31J/mol.K)}$
$\Delta T=24.06K$
Now $\Delta U=\frac{3}{2}nR\Delta T$
We plug in the known values to obtain:
$\Delta U=\frac{3}{2}(3.6mol)(8.31J/mol.K)(24.06K)$
$\Delta U=1079.66J$
(b) We know that
$\Delta T=\frac{2Q}{5nR}$
We plug in the known values to obtain:
$\Delta T=\frac{2(1800J)}{5(3.6mol)(8.31J/mol.K)}$
$\Delta T=24.06K$
(c) We can find the change in the volume of the gas as follows:
$\Delta V=\frac{Q-\Delta U}{P}$
We plug in the known values to obtain:
$\Delta V=\frac{1800J-1079.6J}{120\times 10^3Pa}$
$\Delta V=6.00\times 10^{-3}m^3$
