Question

Nitrogen gas in an expandable container is cooled from 50.0\(^\circ\)C to 10.0\(^\circ\)C with the pressure held constant at 3.00 \(\times\) 10\(^5\) Pa. The total heat liberated by the gas is 2.50 \(\times\) 10\(^4\) J. Assume that the gas may be treated as ideal. Find (a) the number of moles of gas; (b) the change in internal energy of the gas; (c) the work done by the gas. (d) How much heat would be liberated by the gas for the same temperature change if the volume were constant?

Short answer

a) Use ideal gas laws to evaluate mole values. b) Compute \( \Delta U \) from heat and work changes. c) Apply \( W = nR\Delta T \), verifying contracted volumes. d) Utilize \( Q_v \) measurement neglects work via molar heat roles.

Step-by-step solution

Convert Temperatures to Kelvin

Start by converting the given temperatures from degrees Celsius to Kelvin. The conversion formula is \( T(K) = T(^\circ C) + 273.15 \).\(T_1 = 50.0 + 273.15 = 323.15\, K \) \(T_2 = 10.0 + 273.15 = 283.15\, K \)

Calculate Number of Moles (Part a)

Use the ideal gas law \( PV = nRT \) to solve for the number of moles \( n \). Assume the same \( V \) for initial and final conditions due to pressure held constant. Since \( V \) does not change, calculate \( n \) using either condition, for example: \( n = \frac{PV}{RT_1}\)1) Substitute the ideal gas constant \( R = 8.314 \) J/(mol·K): \( n = \frac{3.00 \times 10^5 \times V}{8.314 \times 323.15}\)Here, \( V \) cancels out, and you can rearrange in terms of other known quantities if needed.

Find Change in Internal Energy (Part b)

Use the first law of thermodynamics, \( \Delta U = Q - W \), where \(\Delta U \) is the change in internal energy, \( Q \) is heat change, and \( W \) is work done by the gas. When volume is not constant, solve for \( W \) first via \( W = P\Delta V \), requiring next actual volume calculations if not already asserted constant. With heat liberated, calculate \( \Delta U \): \( \Delta U = Q - W \) Given \( Q = -2.50 \times 10^4 \) J (indicating output), further evaluate with \( W \) understanding.

Calculate Work Done by the Gas (Part c)

Since pressure is constant and volume changes, let's solve for work, \( W = P\Delta V \). Use \(\Delta V = V_2 - V_1 \). Considering the entire reversible process: \( W = nR(T_2 - T_1) \) derived with simplicity is sufficient noting equalities. Given work should be negative indicating cooling, leading to contraction:

Calculate Heat at Constant Volume (Part d)

For constant volume conditions, use: \( Q_v = nC_v \Delta T \) where \( C_v = \frac{f}{2}R \), the constant volume molar heat capacity for diatomic molecules, \( f = 5 \).Replace values for nitrogen: - Compute \( \Delta T = T_2 - T_1 \) using Kelvin: \( Q_v = n \times \left(\frac{5}{2} \times 8.314\right) \times (283.15 - 323.15) \). Recall \( \Delta U \) can also apply due lack of exertion.

Key concepts

First Law of Thermodynamics

The first law of thermodynamics lays the groundwork for understanding energy interactions in physical systems. It's essentially about the conservation of energy, stating that energy cannot be created or destroyed, only transformed or transferred. This law is expressed as: \[ \Delta U = Q - W \]where \( \Delta U \) represents the change in internal energy of the system, \( Q \) is the heat added to the system, and \( W \) is the work done by the system.
In the context of the exercise, as the nitrogen gas cools and contracts under constant pressure, it does work on the surroundings by reducing its volume. The release of heat (\( Q = -2.50 \times 10^4 \) J) results in a decrease in the internal energy of the gas, adjusted by the work term \( W \) due to the change in volume.

• Heat \( Q \) is negative here since the gas is losing heat.
• Work \( W \) signals changes in volume under constant pressure, impacting \( \Delta U \).

Understanding this relationship helps elucidate how transferring energy as heat and work affects a gas's internal state.

Heat Transfer

Heat transfer deals with the movement of thermal energy from a region of higher temperature to a region of lower temperature. In the given exercise, the nitrogen gas undergoes a decrease in temperature, with heat being liberated as it is removed from the gas to its surroundings.
Heat transfer can occur in three ways: conduction, convection, and radiation, but here it is considered in relation to thermodynamic processes happening at constant pressure.

• The process is governed by the ideal gas law, where temperature changes induce volume changes.
• The amount of heat transfer is dictated by the initial and final states of the temperature.

The exercise part (d) adds clarity to how the gas behaves if the volume is held constant. Differentiating between constant volume and constant pressure conditions is crucial. At constant volume, the full heat exchange is used to change the internal energy without performing any work (since \( W = 0 \)).

Molar Heat Capacity

Molar heat capacity is a measure of the amount of heat required to raise the temperature of one mole of a substance by one degree Kelvin. For gases, this concept varies significantly depending on whether the volume or pressure is constant.
When dealing with diatomic gases like nitrogen, different constants are used:

• \( C_p \) (heat capacity at constant pressure): How much heat is needed per mole to heat the gas while allowing it to expand.
• \( C_v \) (heat capacity at constant volume): How much heat is needed when the gas is confined to a fixed volume.

For part (d) of the exercise, it's calculated under constant volume conditions using \( C_v = \frac{f}{2}R \), where \( f = 5 \) for diatomic nitrogen. This indicates that less heat would be liberated with no work done, as all energy change caters to internal energy alterations without any mechanical work exerted.