Question

Consider air to be a diatomic gas with average molecular mass of 29g/mole. A mass 1.45kg of air is contained in a cylinder with a piston at ${27}^{0}C$ and pressure $1.5\times {10}^{5}N/m^2$. Energy is given to the system as heat and the system is allowed to expand till the final pressure $3.5\times {10}^{5}N/m^2$. As the gas expands, pressure and volume follow the relation $V=K{P}^2$ where K is constant (R =8.3J/(mole-K).

Increase of internal energy of the system is nearly :

• A. $3.6\times {10}^{6}J$
• B. $4.2\times {10}^{4}J$
• C. $86\times {10}^{3}J$
• D. $6.3\times {10}^{4}J$

Answer 1

The correct option is A $3.6\times {10}^{6}J$

Given that $V=K{P}^2$

$\therefore \frac{V_1}{{P_1}^2}=\frac{V_2}{{P_2}^2}$

Since, air is assumed as ideal gas,

$V_1=\frac{\left(\frac{1.45\times 1000}{29}\right)\times 8.3\times 300}{1.5\times {10}^5}=0.83m^3$

$V_2=\frac{{(3.5\times {10}^5)}^2}{(1.5\times {10}^5{)}^2}\times 0.83=4.52m^3$

Using ideal gas equation,

$P_2{V}_2=nR{T}_2$

$T_2=\frac{(3.5\times {10}^5)\times 4.52}{\left(\frac{1.45\times 1000}{29}\right)\times 8.3}=3812K$

Change in internal energy, $\Delta U=n{C}_v\Delta T=\left(\frac{1.45\times 1000}{29}\right)\times (\frac{5}{2}\times 8.3)\times (3812-300)=3.6\times {10}^{6}J$

Hence, option A is correct.