Question

What is $\Delta U$ for the process described by figure. Heat supplied during the process $q=100kJ$.

• A. $+50kJ$
• B. $-50kJ$
• C. $-150kJ$
• D. $+150kJ$

Answer 1

The correct option is A $+50kJ$

$q=100kJ$ (given)

$\Delta V=?$

$\Delta U=q+w$ (From first law of thermodynamics)

$w=$ Area under $P-V$ curve

$w=\frac{1}{2}(1+2)\times 1$

$w=\frac{3}{2}$ bar $-m^3$

$w=\frac{3}{2}\times 100k$ $(1bar-m^3=100kJ)$

$w=150kJ$

$\Delta U=-100kJ+150kJ$

$\Delta U=+50kJ$