Question

An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its internal energy increased by 7845 J.
What is the specific heat of the gas?

Answer 1

Applying 1st law of thermodynamics:
Q = Delta U + WB 
WB is the workdone by the system
WB = 346 J 
As,
Delta U = ( m ) ( CV ) ( T2 - T1 ) 

CV = ( Delta U ) / ( m ) ( T2 - T1 ) 
Putting the values in equation:

CV = ( 8485 J ) / ( 80.0 g ) ( 225 C - 25 C ) 

CV = 0.5303 J / g - C degree 
Now putting in original equation:
Q = Delta U + WB 

Q = ( 8485 ) + ( 346 ) = 8831 J
8831J is the specific heat of the gas