Problem 15.54 The p V diagram in the figure ( Figure 1 ) shows a process abc inv
ID: 2256907 • Letter: P
Question
Problem 15.54 The pV diagram in the figure (Figure 1) shows a process abc involving 0.610 of an ideal gas. Part A What was the temperature of this gas at points a, b, and c? Enter your answers separated by commas. Ta, Tb, Tc = 395,6900,1.11 Problem 15.54 The pV diagram in the figure (Figure 1) shows a process abc involving 0.610 of an ideal gas. Problem 15.54 The pV diagram in the figure (Figure 1) shows a process abc involving 0.610 of an ideal gas. The pV diagram in the figure (Figure 1) shows a process abc involving 0.610 of an ideal gas. The pV diagram in the figure (Figure 1) shows a process abc involving 0.610 of an ideal gas. Part A What was the temperature of this gas at points a, b, and c? Enter your answers separated by commas. Ta, Tb, Tc = 395,6900,1.11 Part A What was the temperature of this gas at points a, b, and c? Enter your answers separated by commas. Ta, Tb, Tc = 395,6900,1.11 Part A What was the temperature of this gas at points a, b, and c? Enter your answers separated by commas. Ta, Tb, Tc = 395,6900,1.11 Part A What was the temperature of this gas at points a, b, and c? Enter your answers separated by commas. Ta, Tb, Tc = 395,6900,1.11 Part A What was the temperature of this gas at points a, b, and c? Enter your answers separated by commas. Ta, Tb, Tc = 395,6900,1.11 Part A What was the temperature of this gas at points a, b, and c? Enter your answers separated by commas. Ta, Tb, Tc = 395,6900,1.11 Ta, Tb, Tc = 395,6900,1.11 Ta, Tb, Tc = 395,6900,1.11 395,6900,1.11 Problem 15.54 The pV diagram in the figure (Figure 1) shows a process abc involving 0.610 of an ideal gas. Part A What was the temperature of this gas at points a, b, and c? Enter your answers separated by commas. Ta, Tb, Tc = 395,6900,1.11 The pV diagram in the figure (Figure 1) shows a process abc involving 0.610 of an ideal gas. What was the temperature of this gas at points a, b, and c? How much work was done by or on the gas in this process? How much heat had to be put in during the process to increase the internal energy of the gas by 1.60 times 104J ?Explanation / Answer
A)
by using ideal gas equation
PV=nRT
where,
R=8.3145 J/mol*K
T=PV/nR
at point a
Ta=2*10^5*0.01/0.61*8.3145
Ta=394.333 K.........
and
Tb=5*10^5*0.07/0.61*8.3145
Tb=6900.8418 K .......
and
Tc=8*10^5*0.07/0.61*8.3145
Tc=11041.34688 K .....
B)
workdone dW=area under the curve
=1/2*0.06*3*10^5
=900 J
c)
dQ=dW+dU
=900+16000
=16900 J