one mole of an ideal momoatomic gas at s.t.p undergoes isochoric heating at 25 degrees C. Calculate, the pressure, the heat, the workdone, du, dH.

To calculate the pressure, heat, work done, du (change in internal energy), and dH (change in enthalpy) for an isochoric (constant volume) heating process, we can use the ideal gas law and specific heat formulas.

Given:
Moles of gas = 1 mole
Standard temperature and pressure (STP):
- T = 0 degrees Celsius (273.15 Kelvin)
- P = 1 atm or 101.325 kPa

To calculate the pressure during isochoric heating, we need to use the ideal gas law:

PV = nRT

where:
P = Pressure
V = Volume (constant in this case)
n = Moles of gas
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature

Substituting the values into the ideal gas law, we have:

P * V = n * R * T

Since the volume is constant (isochoric), V * n = constant, so:

P = (constant) * T

At STP, the volume of 1 mole of an ideal gas is about 22.4 L, so:

P = (22.4 L) * (1 atm) / (273.15 K) * (25 + 273.15 K)
P ≈ 0.0821 * 298.15

Therefore, the pressure is approximately 24.56 atm.

Next, to calculate the heat (Q), we can use the equation:

Q = n * C_v * ΔT

where:
Q = Heat
n = Moles of gas
C_v = Molar heat capacity at constant volume
ΔT = Change in temperature

For an ideal monoatomic gas, the molar heat capacity at constant volume (C_v) is calculated as:

C_v = (3/2) * R

Substituting the values, we have:

Q = (1 mol) * (3/2 * 0.0821 L·atm/mol·K) * (25 - 0) K
Q = 3.07575 L·atm

Therefore, the heat is approximately 3.08 L·atm.

Since the process is isochoric (constant volume), no work is done (W = 0) because work is defined by the equation:

W = P * ΔV

where ΔV is the change in volume, which is zero for an isochoric process.

The change in internal energy (ΔU) and change in enthalpy (ΔH) for an ideal gas can be calculated using the following equations:

ΔU = Q - W
ΔH = ΔU + P * ΔV

Since W = 0, the change in internal energy (ΔU) is equal to the heat (Q):

ΔU = Q = 3.08 L·atm

Therefore, the change in internal energy is approximately 3.08 L·atm.

To find the change in enthalpy (ΔH), we need to consider the equation:

ΔH = ΔU + P * ΔV

Since the process is isochoric (ΔV = 0), the change in enthalpy is equal to the change in internal energy:

ΔH = ΔU = 3.08 L·atm

Therefore, the change in enthalpy is approximately 3.08 L·atm.

To summarize:
- Pressure (P) ≈ 24.56 atm
- Heat (Q) ≈ 3.08 L·atm
- Work done (W) = 0
- Change in internal energy (ΔU) ≈ 3.08 L·atm
- Change in enthalpy (ΔH) ≈ 3.08 L·atm