Question

1) 15 moles of oxygen are at the initial state with pressure p = 10^5 Pa and volume V = 0.35 m^3. The gas expands quasistatically to V = 0.70 m^3. What is its final equilibrium temperature if the process is adiabatic? A. T= 102.4 K B. 175.2 K C. 212.8 K D. 305.7 K 2) Calculate the heat Q produced during the isobaric cooling process of ideal gas sample with adiabatic coefficient ? of 1.4, if the work W done by the compression is 16 J. A. Q= 56 J B. Q= 72 J C. Q= 12 J D. Q= 25 J 

          1) 15 moles of oxygen are at the initial state with pressure p = 10^5 Pa and volume V = 0.35 m^3. The gas expands quasistatically to V = 0.70 m^3. What is its final equilibrium temperature if the process is adiabatic?
A. T= 102.4 K B. 175.2 K C. 212.8 K D. 305.7 K

2) Calculate the heat Q produced during the isobaric cooling process of ideal gas sample with adiabatic coefficient ? of 1.4, if the work W done by the compression is 16 J.
A. Q= 56 J B. Q= 72 J C. Q= 12 J D. Q= 25 J
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1) 15 moles of oxygen are at the initial state with pressure p = 10^5 Pa and volume V = 0.35 m^3. The gas expands quasistatically to V = 0.70 m^3. What is its final equilibrium temperature if the process is adiabatic? A. T= 102.4 K B. 175.2 K C. 212.8 K D. 305.7 K 2) Calculate the heat Q produced during the isobaric cooling process of ideal gas sample with adiabatic coefficient ? of 1.4, if the work W done by the compression is 16 J. A. Q= 56 J B. Q= 72 J C. Q= 12 J D. Q= 25 J

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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1) 15 moles of oxygen are at the initial state with pressure p = 10^5 Pa and volume V = 0.35 m^3. The gas expands quasistatically to V = 0.70 m^3. What is its final equilibrium temperature if the process is adiabatic? A. T = 102.4 K B. 175.2 K C. 212.8 K D. 305.7 K 2) Calculate the heat Q produced during the isobaric cooling process of ideal gas sample with adiabatic coefficient ̲ of 1.4, if the work W done by the compression is 16 J. A. Q = 56 J B. Q = 72 J C. Q = 12 J D. Q = 25 J
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Transcript

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00:01 Hi, let's see for adiabetic process we have t v power y minus 1 equal to constant.
00:09 So t1 v1 power y minus 1 equal to t2 v2 power y minus 1.
00:18 So for t 1 we're going to use p1 into v1 equal to n r t 1.
00:26 So t1 equal to p1 v1 by nr.
00:32 Let's substitute the value here.
00:34 So t1 equal to 10 to the power 5 into 0 .35 divided by 15 into 8 .314.
00:46 That is equal to 280 .65 kelvin.
00:51 So we can write the above equation as 280 .6...
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