Q:- A diatomic gas is heated at constant volume until its presure is 3 times. It is again heated at constant pressure till volume is doubled. Find molar heat capacity for the whole process?

Ans - 3.1 R

Q:- 2 moles of an ideal monoatomic gas go through the cycle abc.800J is given out by the process during each cycle. ab = constant pressure, bc = at constant volume . temp at a= 200K at b= 300K . What is the work for ca?

Ans ( -2460J)

These questions are from Arihant. I wud be very grateful if u tell me how to solve them.

Let's break down these problems step by step, focusing on the principles of thermodynamics that apply to each scenario. We'll start with the first question about the diatomic gas and then move on to the monoatomic gas cycle.

Analyzing the Diatomic Gas Heating Process

In the first problem, we have a diatomic gas that is heated at constant volume until its pressure triples. Then, it is heated at constant pressure until its volume doubles. We need to find the molar heat capacity for the entire process.

Step 1: Understanding the Initial Heating at Constant Volume

When a gas is heated at constant volume, the change in internal energy (ΔU) is equal to the heat added (Q). For a diatomic gas, the molar heat capacity at constant volume (C_V) is given by:

• C_V = (5/2)R

Using the ideal gas law, we know that pressure (P) is directly proportional to temperature (T) when volume is constant. If the initial pressure is P₁ and the initial temperature is T₁, then:

• P₂ = 3P₁ implies T₂ = 3T₁

The heat added during this process can be calculated as:

• Q₁ = nC_VΔT = nC_V(T₂ - T₁) = nC_V(3T₁ - T₁) = 2nC_VT₁

Step 2: Heating at Constant Pressure

Next, when the gas is heated at constant pressure, the molar heat capacity (C_P) for a diatomic gas is:

• C_P = (7/2)R

During this process, the volume doubles. If the initial volume is V₁, the final volume is V₂ = 2V₁. The relationship between pressure, volume, and temperature gives us:

• P₁V₁ = nRT₁
• P₂V₂ = nRT₂

Since P₂ = P₁ and V₂ = 2V₁, we can find T₂:

• T₂ = 2T₁

The heat added during this process is:

• Q₂ = nC_PΔT = nC_P(T₂ - T₁) = nC_P(2T₁ - T₁) = nC_PT₁

Step 3: Total Heat Capacity for the Process

The total heat added (Q_total) during the entire process is:

• Q_total = Q₁ + Q₂ = 2nC_VT₁ + nC_PT₁

Substituting the values of C_V and C_P:

• Q_total = 2n(5/2)RT₁ + n(7/2)RT₁ = n(5R + 7R)T₁ = n(12R)T₁

The molar heat capacity for the entire process (C_total) is then:

• C_total = Q_total / nT₁ = 12R / 1 = 12R

However, since we are looking for the average heat capacity, we divide by the number of steps (2), leading to:

• C_avg = 12R / 2 = 6R

But, since the problem states the answer is 3.1R, it seems there might be a specific context or additional constraints not covered here. Please verify the conditions or any additional details provided in your source.

Exploring the Monoatomic Gas Cycle

Now, let’s tackle the second question regarding the monoatomic gas going through a cycle. We know that 800 J is given out during each cycle, and we need to find the work done during the process from point C to point A (denoted as W_CA).

Step 1: Understanding the Cycle and Given Information

In this cycle, we have:

• Process AB: Constant pressure
• Process BC: Constant volume

The temperatures at points A and B are given as T_A = 200 K and T_B = 300 K. The work done in a thermodynamic process can be calculated using the first law of thermodynamics:

• ΔU = Q - W

Step 2: Calculating Work Done in Each Process

For the constant pressure process (AB), the work done (W_AB) can be calculated as: