Volume change of ideal gas

Classical Mechanics Level 1

The above is the pressure-temperature diagram of a given mass of ideal gas which is circulating by the sequence A $\to$ B $\to$ C $\to$ A. Let V $_A$ , V $_B$ and V $_C$ denote the volumes at A, B and C, respectively, then what is the ratio $V_A:V_B:V_C?$

1 : 1 : 3 3 : 2 : 4 2 : 3 : 6 3 : 3 : 1

4 solutions

Siva Chowdary
Mar 12, 2014

pv/t=constant

Jung Hyun Ran
May 19, 2014

$\frac { P\times V }{ T } =constant\quad therefore\\ \\ A\longrightarrow B:\quad \\ \frac { { P }_{ A }\times { V }_{ A } }{ { T }_{ 0 } } =\frac { { P }_{ B }\times { V }_{ B } }{ { T }_{ 0 } } \\ \frac { { P }_{ 0 }\times { V }_{ A } }{ { T }_{ 0 } } =\frac { { 2P }_{ 0 }\times { V }_{ B } }{ { 3T }_{ 0 } } \\ \Rightarrow { V }_{ A }=\frac { 2 }{ 3 } { V }_{ B }\quad \quad \quad \left( 1 \right) \\ \\ B\longrightarrow C:\quad \\ \frac { { P }_{ B }{ \times V }_{ B } }{ { T }_{ B } } =\frac { { P }_{ C }\times { V }_{ C } }{ { T }_{ C } } \\ \frac { { 2P }_{ 0 }\times { V }_{ B } }{ { 3T }_{ 0 } } =\frac { { P }_{ 0 }\times { V }_{ C } }{ { 3T }_{ 0 } } \\ { 2P }_{ 0 }\times { V }_{ B }={ P }_{ 0 }\times { V }_{ C }\quad \\ \Rightarrow { V }_{ B }=\frac { 1 }{ 2 } { V }_{ C }\quad \quad \quad \left( 2 \right) \\ \\ C\longrightarrow A:\quad \\ \frac { { P }_{ C }\times { V }_{ C } }{ { T }_{ C } } =\frac { { P }_{ A }\times { V }_{ A } }{ { T }_{ A } } \\ \frac { { P }_{ 0 }\times { V }_{ C } }{ { 3T }_{ 0 } } =\frac { { P }_{ 0 }\times { V }_{ A } }{ { T }_{ 0 } } \\ \Rightarrow { V }_{ C }=3{ V }_{ A }\quad \quad \quad \left( 3 \right) \\ \\ \left( 1 \right) \left( 2 \right) \left( 3 \right) \Longrightarrow { V }_{ A }{ :V }_{ B }{ :V }_{ C }=2:3:6\\ \\ \\$

Yadubansh Mani
Feb 26, 2014

PV/T=constant Hence the ratio is 2:3:6

Andre Yudhistika
Feb 26, 2014

we know that pv/T=constant therefore

v=T/p

v1=1 , v2=3/2 , v3=3

v1:v2:v3

1:3/2:3 x2

2:3:6

Volume change of ideal gas

4 solutions

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