Question

The pressure of a monoatomic gas increases linearly from $4\times {10}^5{Nm}^{-2}to8\times {10}^{-5}{Nm}^{-2}$ when its volume increases from ${0.2m}^3$ to ${0.5m}^3$. calculate (a) work done by the gas, (b) increase in internal energy (c) mass of heat supplied and (d) molar heat capacity of the gas.

Answer 1

$P_1=4\times {10}^{5}N{m}^{-2}{V}_1=0.2m^3{P}_2=8\times {10}^{-5}N{m}^{-2}{V}_2=0.5m^3$

$a)$ Work done = Area under P-V graph

area(ABCDEA)

= area of $△ABC$ + area of rectangle $ACDE$

$=\frac{1}{2}(BC\times AC)+area\left(ACDE\right)=\frac{1}{2}(4\times {10}^5\times 0.3)+(0.3\times 4\times {10}^5)=180000J=1.8\times {10}^{5}J$

$b)$ Since it is a monoatomic gas $\Rightarrow C_V=\frac{3}{2}R$

Increase in internal energy :

$△U=[C_V(T_2-T_1)-\frac{C_V}{R}R(T_2-T_1)+\frac{C_V}{R}(P_2{V}_2-P_1{V}_1)]△U=\frac{C_V}{R}(P_2{V}_2-P_1{V}_1)=\frac{3}{2}\left[(8\times {10}^5)\right(0.5)-(4\times {10}^5)(0.2\left)\right]=4.8\times {10}^{5}J$

$c)$ Heat supplied $\theta =△U+W=4.8\times {10}^5+1.8\times {10}^5=6.6\times {10}^{5}J$

$d)$ Molar heat capacity

$C=\frac{\theta }{△T}=\frac{\theta R}{R△T}=\frac{\theta R}{(P_2{V}_2-P_1{V}_1)}=\frac{6.6\times {10}^5\times 8.31}{\left[(8\times {10}^5)\right(0.5)-(4\times {10}^5)(0.2\left)\right]}=17.14J{mol}^{-1}{K}^{-1}$