A monatomic ideal gas is compressed adiabatically from a pressure of 1.09 105 Pa and volume of 250 L to a volume of 39.0 L.

(a) What is the new pressure of the gas?
(b) How much work is done on the gas?

P1*V1=P2*V2

1 liter= .001 m^3

To find the new pressure of the gas, we can use the relationship between pressure and volume for an adiabatic process. The equation is given by:

P1 * V1^γ = P2 * V2^γ

where P1 and V1 are the initial pressure and volume, P2 and V2 are the final pressure and volume, and γ is the heat capacity ratio.

(a) Let's calculate the new pressure of the gas:

Given:
Initial pressure (P1) = 1.09 x 10^5 Pa
Initial volume (V1) = 250 L
Final volume (V2) = 39.0 L
Heat capacity ratio (γ) for monatomic ideal gas = 5/3

Using the formula:

P1 * V1^γ = P2 * V2^γ

P2 = (P1 * V1^γ) / V2^γ
P2 = (1.09 x 10^5 Pa * (250 L)^(5/3)) / (39.0 L)^(5/3)

Calculating this expression will give us the new pressure of the gas.

(b) The work done on the gas can be determined using the equation:

W = ∆U + ∆Q

Since the process is adiabatic (no heat transfer), ∆Q = 0. Therefore, the equation simplifies to:

W = ∆U

The change in internal energy (∆U) can be calculated using the first law of thermodynamics for an ideal gas:

∆U = Q - W

Since ∆Q = 0, the equation becomes:

W = -∆U

Here, we need to calculate the change in internal energy (∆U) to determine the work done on the gas.

Please provide the necessary values to calculate the new pressure and the work done on the gas.