Question

1. Under constant-volume conditions, 4200 J of heat is added to 1.4 moles of an ideal gas. As a result, the temperature of the gas increases by 103 K. How much heat would be required to cause the same temperature change under constant-pressure conditions? Do not assume anything about whether the gas is monatomic, diatomic, etc.

2. A system gains 3080 J of heat at a constant pressure of 1.36 × 105 Pa, and its internal energy increases by 4160 J. What is the change in the volume of the system, and is it an increase or a decrease? ΔV = Entry field with correct answer Entry field with incorrect answer Entry field with correct answer

3. A system undergoes a two-step process. In the first step, the internal energy of the system increases by 229 J when 166 J of work is done on the system. In the second step, the internal energy of the system increases by 115 J when 193 J of work is done on the system. For the overall process, find the heat added to the system (including sign).

Answer 1

Solution (1):

At constant volume, heat added in a gas is given by

     (1)

At constant pressure, heat added in a gas is given by

   (2)

Given that:

Putting the value of from equation (1), in (2), we find

(3)

So, for same increase in temperature, the heat added in constant Pressure Process is times of head added in constant volume Process.