Question

An ideal gas is enclosed in a cylinder with a movable piston on top. The piston has mass of $8000gm$ and an area of $5.00c{m}^2$ and is free to slide up and down, Keeping the pressure of the gas constant. How much work is done (in $J$) as the temperature of $0.200mol$ of the gas is raised from ${200}^{o}Cto{300}^{o}C$

Answer 1

The first law of thermodynamics gives us the relation:
$\delta q=\Delta U+w$
$\delta q=\int nCdT$ where C is the specific heat capacity for the process and n is the no of moles of the ideal gas.
Here, the process is a constant pressure process, hence $C=C_p$
And the change in internal energy for an ideal gas is given by $\int n{C}_{v}dT$.
i.e.
$\int n{C}_{p}dT=\int n{C}_{v}dT+w$
$\int n(C_p-C_v)dT=w$ ; $C_p-C_v=R$ for an ideal; R (universal gas constant)
i.e. $w=\int nRdT=nR(T_2-T_1)=0.2\times 8.314\times (300-200)=0.2\times 8.314\times 100=166J$