`4.0 g` of a gas occupies `22.4` litres at NTP. The specific heat capacity of the gas at constant volume is `5.0 JK^(-1)mol^(-1)`. If the speed of sound in this gas at NTP is `952 ms^(-1)`. Then the heat capacity at constant pressure is

Options

A. `8.5 JK^(-1) mol^(-1)`
B. `8.0 JK^(-1) mol^(-1)`
C. `7.5 JK^(-1) mol^(-1)`
D. `7.0 JK^(-1) mol^(-1)`

Explanation

To find the heat capacity at constant pressure (\(C_P\)) for the given gas, we can use the relationship between the speed of sound in a gas and its specific heat capacities. The formula relating the speed of sound (\(V\)), specific heat capacities (\(C_P\) and \(C_V\)), and the gas constant (\(R\)) is given by:

\[ V = \sqrt{\frac{\gamma R T}{M}} \]

where:
- \(\gamma = \frac{C_P}{C_V}\)
- \(R\) is the universal gas constant, approximately \(8.314 \, \text{J K}^{-1} \text{mol}^{-1}\)
- \(T\) is the temperature in Kelvin
- \(M\) is the molar mass of the gas

### Step 1: Determine the Molar Mass of the Gas
Given that \(4.0 \, \text{g}\) of the gas occupies \(22.4 \, \text{L}\) at NTP (Normal Temperature and Pressure), we can find the molar mass (\(M\)) of the gas.

At NTP, \(1 \, \text{mol}\) of any ideal gas occupies \(22.4 \, \text{L}\). Therefore, the number of moles (\(n\)) of the gas can be calculated as:

\[ n = \frac{\text{mass}}{\text{molar mass}} \Rightarrow n = \frac{4.0 \, \text{g}}{M} \]

Since \(4.0 \, \text{g}\) occupies \(22.4 \, \text{L}\), we have:

\[ \frac{4.0 \, \text{g}}{M} = 1 \, \text{mol} \Rightarrow M = 4.0 \, \text{g/mol} \]

### Step 2: Calculate \(\gamma\)
We know \(C_V = 5.0 \, \text{J K}^{-1} \text{mol}^{-1}\). Using the relationship:

\[ \gamma = \frac{C_P}{C_V} \]

We can express \(C_P\) in terms of \(\gamma\):

\[ C_P = \gamma C_V \]

### Step 3: Substitute into the Speed of Sound Equation
From the speed of sound equation, we can express \(\gamma\):

\[ V = 952 \, \text{m/s} = \sqrt{\frac{\gamma \cdot R \cdot T}{M}} \]

Substituting the known values:

- \(R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1}\)
- \(T = 273 \, \text{K}\)
- \(M = 0.004 \, \text{kg/mol}\) (since \(4.0 \, \text{g} = 0.004 \, \text{kg}\))

### Step 4: Solve for \(\gamma\)
Squaring both sides gives:

\[ 952^2 = \frac{\gamma \cdot 8.314 \cdot 273}{0.004} \]

Calculating the left side:

\[ 952^2 = 906304 \]

Now, rearranging for \(\gamma\):

\[ \gamma = \frac{906304 \cdot 0.004}{8.314 \cdot 273} \]

Calculating the denominator:

\[ 8.314 \cdot 273 \approx 2270.862 \]

Now substituting back:

\[ \gamma = \frac{3625.216}{2270.862} \approx 1.597 \]

### Step 5: Calculate \(C_P\)
Now substituting \(\gamma\) back into the equation for \(C_P\):

\[ C_P = \gamma C_V = 1.597 \cdot 5.0 \approx 7.985 \approx 8.0 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Final Answer
Thus, the heat capacity at constant pressure \(C_P\) is approximately:

\[ \boxed{8.0 \, \text{J K}^{-1} \text{mol}^{-1}} \]