a gas occupies 2 m3 at 100k and exerts a pressure of 100kpa. What volume will the gas occupy if the temperature is increased to 400.0k and the pressure is increased to 200.0kPa?

PV = nRT

Make sure P, V, R units are consistent.

To solve this problem, we can use the ideal gas law equation, which states:

PV = nRT

Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant
T = temperature of the gas

In this case, we are given the initial volume (V1 = 2 m³), initial pressure (P1 = 100 kPa), and the final temperature (T2 = 400 K) and final pressure (P2 = 200 kPa). We need to find the final volume (V2).

First, we can rearrange the ideal gas law equation to solve for V2:

V2 = (P1 * V1 * T2) / (P2 * T1)

Now, let's plug in the values into the equation:

V2 = (100 kPa * 2 m³ * 400 K) / (200 kPa * 100 K)

Simplified, we get:

V2 = (200 kPa * m³ * K) / (200 kPa)

The kPa units cancel out, leaving us with:

V2 = 2 m³

Therefore, the gas will occupy 2 m³ of volume when the temperature is increased to 400.0 K and the pressure is increased to 200.0 kPa.