A process `1rightarrow2` using monoatomic gas is shown on the P-V diagram on the right `P_(1)=2P_(2)=10^(6)N//m^(2),V_(2)=4V_(1)=0.4m^(3)` The heat ansorbed by the gas

Options

A. 350kJ
B. 375kJ
C. 425kJ
D. none

Answer

B

Explanation

`W=1/2xx3v_(0)xxP_(0)+3v_(0)xxP_(0)`
`W=3/2P_(0)v_(0)+3v_(0)P_(0)`
`W=a/2P_(0)v_(0)`
`DeltaU=nC_(v)DeltaT=n((3R)/2)(T_(f)-T_(i))`
`DeltaU= 3/2nR (T_(f)-T_(i))=3/2[P_(f)V_(f)-P_(i)V_(i)]`
`DeltaU-3/2[4P_(0)V_(0)-2P_(0)V_(0)]`
`DeltaU=3P_(0)V_(0)`
`DeltaQ=DeltaU+W=9/2P_(0)V_(0)+3P_(0)V_(0)=15/2P_(0)V_(0)`
`P_(0)=10^(6)/2 , V_(0)=0.1`
`DeltaQ=15/2xx10^(6)/2xx0.1=375000J`
`DeltaQ=375kJ`