Question

61. 1 mol of monoatomic ideal gas perform the
following operations (All steps are reversible) and
pressure at point A is 1 atm
The pressure at point B is ? ​

Answer 1

4

Is it helpful for you??

Then pick me as brainliest

I will be grateful for you?

Answer 2

The pressure at point B is 2.29 atm

Pressure at point A = 1 atm

From the graph, we get,

Temperature at point A = 75 K

Temperature at point B = 300 K

Now, the ratio of the specific gas constant of a mono atomic gas (γ) = 5/3

According to the adiabatic process,

P$V^{\gamma}$ = constant where P is the pressure and V is the volume.

From this equation, we get,

P$T^{\frac{\gamma }{1- \gamma} }$ = constant

By applying this formula we get,

$\frac{P_{A}}{P_{B}}$ = $(\frac{T_{B}}{T_{A}} )^{\frac{\gamma }{1- \gamma } }$

or, $\frac{P_{A}}{P_{B}}$ = $(\frac{300}{75} )^{-\frac{5}{2} }$

or, $P_{B}$ = 1 × 2.29

          = 2.29 atm

The pressure at point B is 2.29 atm