Given:
- Mass of gas, m = 0.5 kg
- Specific heat at constant volume, cv = 0.95 kJ/kgK
- Height from which the box falls, h = 4 km

To find:
- Increase in temperature of the gas when it hits the ground

Assumptions:
- The box is well-insulated, so no heat exchange occurs with the surroundings during the fall
- The gas behaves like an ideal gas

Solution:

1. Determining the initial potential energy of the box:

Potential energy = mgh
where g is the acceleration due to gravity, and h is the height from which the box falls

g = 9.8 m/s^2 (approximate value)
h = 4 km = 4000 m

Potential energy = 0.5 x 9.8 x 4000 = 19,600 J

2. Converting the initial potential energy to internal energy of the gas:

When the box falls, its potential energy is converted to kinetic energy, which is transferred to the gas molecules as internal energy.

ΔU = Q = 19,600 J (where ΔU is the change in internal energy, and Q is the heat transferred)

3. Finding the temperature rise of the gas:

ΔU = mcΔT
where c is the specific heat at constant volume

ΔT = ΔU / mc

Substituting the given values,

ΔT = 19,600 / (0.5 x 0.95)
= 41,052 K

However, this result is not physically meaningful, as it implies a very high temperature rise. This is because the assumption of ideal gas behavior breaks down at such high temperatures.

4. Applying the ideal gas law to estimate the final temperature:

PV = mRT
where P is the pressure, V is the volume, R is the gas constant, and T is the temperature

Assuming that the pressure and volume of the gas remain constant during the fall (since the box is well-insulated),

T = (PV) / (mR)

Using the ideal gas law,

P = ρgh
where ρ is the density of air (taken as 1.2 kg/m^3)

Substituting the given values,

T = [(1.2 x 9.8 x 4000) / 0.5] / (0.287)
= 33,154 K

5. Finding the temperature rise of the gas:

The initial temperature of the gas is assumed to be the same as the ambient temperature at the starting height, which is around -15°C (or 258 K).

Therefore,

ΔT = 33,154 - 258
= 32,896 K

However, this result is also not physically meaningful, as it implies a much higher temperature rise than what the gas can actually withstand.

6. Adjusting the final temperature using the specific heat:

Since the specific heat at constant volume is given, we can use it to estimate a more reasonable temperature rise.

ΔU = mcΔT

Substituting the given values,

19,600 = 0.5 x 0.95 x ΔT

ΔT = 41,052