If Pd vs. P(where P denotes pressure in atm and d denotes density in gm/L ) is plotted for He gas (assume ideal ) at a particular temperature. If `[(d)/(dP)(pd)]_(P=8.21"atm")=5,` then the temperature wil be

Options

A. 160 K
B. 320 K
C. 80 K
D. none of these

Explanation

To solve the problem step by step, we will use the relationship between pressure, density, and temperature for an ideal gas.

### Step 1: Understand the relationship between density, pressure, and temperature.
For an ideal gas, the density \( d \) can be expressed as:
\[ d = \frac{PM}{RT} \]
where:
- \( P \) = pressure in atm,
- \( M \) = molar mass of the gas (for Helium, \( M = 4 \, \text{g/mol} \)),
- \( R \) = ideal gas constant \( = 0.0821 \, \text{L atm/(K mol)} \),
- \( T \) = temperature in Kelvin.

### Step 2: Multiply both sides by pressure \( P \).
We can rewrite the equation as:
\[ Pd = \frac{P^2 M}{RT} \]

### Step 3: Differentiate \( Pd \) with respect to \( P \).
To find \( \frac{d(Pd)}{dP} \), we differentiate:
\[ \frac{d(Pd)}{dP} = \frac{d}{dP}\left(\frac{P^2 M}{RT}\right) = \frac{2PM}{RT} \]

### Step 4: Substitute the given values.
We know from the problem that:
\[ \left[\frac{d(Pd)}{dP}\right]_{P=8.21 \, \text{atm}} = 5 \]
Thus, we can set up the equation:
\[ \frac{2PM}{RT} = 5 \]
Substituting \( P = 8.21 \, \text{atm} \) and \( M = 4 \, \text{g/mol} \):
\[ \frac{2 \times 8.21 \times 4}{RT} = 5 \]

### Step 5: Solve for \( T \).
Rearranging the equation gives:
\[ RT = \frac{2 \times 8.21 \times 4}{5} \]
Calculating the right side:
\[ RT = \frac{65.68}{5} = 13.136 \]

Now, substituting \( R = 0.0821 \, \text{L atm/(K mol)} \):
\[ T = \frac{13.136}{0.0821} \approx 160 \, \text{K} \]

### Final Answer:
The temperature \( T \) is approximately **160 K**.

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