The specific heats of argon at constant pressure and constant volume are `525 J//kg` and `315 J//kg`, respectively. Its density at `NTP` will be\

Options

A. `0.64kg//m^(3)`
B. `2.62kg//m^(3)`
C. `1.20kg//m^(3)`
D. `1.75kg//m^(3)`

Answer

To find the density of argon at Normal Temperature and Pressure (NTP), we can follow these steps:

### Step 1: Identify Given Values
- Specific heat at constant pressure, \( C_P = 525 \, \text{J/kg K} \)
- Specific heat at constant volume, \( C_V = 315 \, \text{J/kg K} \)
- Pressure at NTP, \( P = 10^5 \, \text{Pa} \)
- Temperature at NTP, \( T = 293 \, \text{K} \)

### Step 2: Use the Relationship Between \( C_P \), \( C_V \), and the Gas Constant \( R \)
We know that:
\[ C_P - C_V = R \]
However, since we are dealing with specific heats per unit mass, we can express this as:
\[ C_P - C_V = \frac{R}{M} \]
where \( M \) is the molar mass of argon.

### Step 3: Calculate \( R \) in Terms of Specific Heats
From the equation above, we can rearrange it to find \( R \):
\[ R = M (C_P - C_V) \]

### Step 4: Use the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
where \( n \) is the number of moles. We can express \( n \) as:
\[ n = \frac{m}{M} \]
Substituting this into the ideal gas law gives:
\[ PV = \frac{m}{M} RT \]
Rearranging this, we find:
\[ \frac{R}{M} = \frac{PV}{mT} \]

### Step 5: Relate Density to Pressure and Temperature
We can express density \( \rho \) as:
\[ \rho = \frac{m}{V} \]
Substituting this into the equation gives:
\[ \frac{R}{M} = \frac{P}{\rho T} \]
Rearranging this leads to:
\[ \rho = \frac{P}{T} (C_P - C_V) \]

### Step 6: Substitute Known Values
Now we can substitute the known values into the equation:
\[ \rho = \frac{10^5 \, \text{Pa}}{293 \, \text{K}} \times (525 \, \text{J/kg K} - 315 \, \text{J/kg K}) \]
Calculating \( C_P - C_V \):
\[ C_P - C_V = 525 - 315 = 210 \, \text{J/kg K} \]
Now substituting this back into the density equation:
\[ \rho = \frac{10^5}{293} \times 210 \]

### Step 7: Calculate the Density
Calculating the right-hand side:
\[ \rho = \frac{10^5 \times 210}{293} \approx 1.62 \, \text{kg/m}^3 \]

### Final Answer
Thus, the density of argon at NTP is approximately:
\[ \rho \approx 1.62 \, \text{kg/m}^3 \]

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