Question

Find the total change in the internal energy of a gas that is subjected to the following two-step process. In the first step the gas is made to go through isochoric heating until 5560 J of heat is transferred into the gas and its pressure is 3.32 ✕ 105 Pa. In the second step it is subjected to isobaric compression until its volume decreases by 7.30 ✕ 10−3 m3 and 1270 J of heat is transferred out of the gas. What is the total change in internal energy of this gas

Answer 1

Answer:

U = 9253.6 J

Explanation:

It is given that,

Heat transferred in isochoric process, $Q_1=5560\ J$

Pressure in this process, $P_1=3.32\times 10^5\ Pa$

In the second step,

It is subjected to isobaric compression until its volume decreases by, $V=-7.3\times 10^{-3}\ m^3$ (it decreases)

Heat transferred out of the gas, $Q_2=1270\ J$

In isochoric process, work done by the gas is zero as in this type volume is constant, $W_1=0$

According to first law of thermodynamics,

$\Delta U=Q-W$

$\Delta U_1=5560\ J$.............(1)

In isochoric compression, $W_2=P\Delta V$

$\Delta U_2=1270-3.32\times 10^5\times (-7.3\times 10^{-3})$

$\Delta U_2=3693.6\ J$

Let U is the total change in the internal energy of this gas. So,

$U=U_1+U_2$

$U=5560+3693.6$

U = 9253.6 J

So, the total change in the internal energy of this gas is $$9253.6 J. Hence, this is the required solution.