Amixture of helium and argon gas is expanded from a volume of 29.0l to a volume of 82.0l, while the pressure is held constant at 60.0atm. calculate the work done on the gas mixture. round your answer to 3 significant digits, and be sure it has the correct sign (positive or negative)

322 kJ

Explanation:

The work is the energy that a force produces when realizes a displacement. So, for a gas, it occurs when it expands or when it compress.

When the gas expands it realizes work, so the work is positive, when it compress, it's suffering work, so the work is negative.

For a constant pressure, the work can be calcutated by:

W = pxΔV, where W is the work, p is the pressure, and ΔV is the volume variation. To find the work in Joules, the pressure must be in Pascal (1 atm = 101325 Pa), and the volume in m³ (1 L = 0.001 m³), so:

p = 60 atm = 6.08x10⁶ Pa

ΔV = 82.0 - 29.0 = 53 L = 0.053 m³

W = 6.08x10⁶x0.053

W = 322x10³ J

W = 322 kJ

the first, second, and fifth options

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.