Question

One mole of an ideal gas is put through a series of changes as shown in the figure in which 1,2,3 mark the three stages of the system. Find the Pressure at the three stages.

• A. 1] - 1.107, 2] - 2.214, 3] - 1.107
• B. 1] - 1.113,2] - 2.226, 3] - 2.226
• C. 1] - 2.214,2] - 1.107, 3] - 2.214
• D. 1] - 1.107, 2] - 2.214, 3] - 2.214

Answer 1

Answer: (B)

At stage (1)

$V=22.4L=22.4\times 10^{-3}{m}^3$

$T=300K,n=1mol$

$p=\frac{nRT}{V}=\frac{1mol\times 8.314Pa{m}^{3}mo{l}^{-1}{K}^{-1}\times 300K}{22.4\times 10^{-3}{m}^3}$

$=1.113\times 10^{5}Pa=1.113bar$

At stage (2) temperature is doubled but volume is constant. Thus, $P\propto T,=2.226$ bar

At stage $(3),T$ is $300K$ and $V$ is $11.2L$ and as in stage (1)

$p=\frac{nRT}{V}=2.226bar$