Aballoon filled with helium gas at 20°c occupies 4.91 l at 1.00 atm. the balloon is immersed in liquid nitrogen at -196°c, while the pressure is raised to 5.20 atm. what is the volume of the balloon in the liquid nitrogen?

V = 0.248 L

Explanation:

To do this, use the following equation:

P1*V1/T1 = P2*V2/T2

This equation is used to find a relation between two differents conditions of a same gas, which is this case. From this equation we can solve for V2.

Solving for V2:

V2 = P1*V1*T2/T1*P2

Temperature must be at Kelvin, so, we have to sum the temperature 273 to convert it in K.

Replacing the data we have:

V2 = 1 * 4.91 * (-196+273) / 5.2 * (20+273)

V2 = 378.07 / 1523.6

V2 = 0.248 L

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.