Answer
(a) $0.05m^3$
(b) $-1.22KJ$
(c) $-1.22KJ$
Work Step by Step
(a) We know that
$V_f=\frac{nRT}{P_f}$
We plug in the known values to obtain:
$V_f=\frac{(2.75mol)(8.31J/mol.K)(295K)}{120\times 10^3Pa}$
$V_f=0.05m^3$
(b) The required work done can be calculated as
$W=nRT\ln(\frac{P_i}{P_f})$
We plug in the known values to obtain:
$W=(2.75mol)(8.31J/mol.K)(295K)\ln(\frac{101\times 10^3Pa}{120\times 10^3Pa})$
$W=-1.22KJ$
(c) We know that
$\Delta Q=\Delta U+W$
We plug in the known values to obtain:
$\Delta Q=0-1.22KJ$
$\Delta Q=-1.22KJ$
