Question

If $C_v=4.96cal/mol.K$ then the amount of heat absorbed at constant volume, when temperature of 2 moles of this gas is increased from $332K$ to $542K$ is:

• A. $5208cal$
• B. $2083cal$
• C. $8688cal$
• D. $3256cal$

Answer 1

The correct option is B $2083cal$

Change in internal energy is equal to heat supplied at constant volume i.e ,
$△U=q_v=n{C}_v△T$
$n=2moles$
$T_2=542K$
$T_1=332K$
$C_v=4.96cal/mol.K$
$q_v=2\times 4.96\times (542-332)=2083cal$