50g of oxygen at NTP is compressed adiabatically to a pressure of 5 atmosphere. The work done on the gas, if `gamma =1.4 and R = 8.31 J mol^(-1)K^(-1)` is

Options

A. `-5173 J`
B. `1131 J`
C. `-1364 J`
D. `5673 J`

Explanation

To solve the problem, we need to calculate the work done on 50g of oxygen gas during an adiabatic compression from NTP to a pressure of 5 atm. We will follow these steps:

### Step 1: Calculate the number of moles of oxygen (n)
The molecular mass of oxygen (O₂) is 32 g/mol (16 g/mol for each oxygen atom).

\[ n = \frac{\text{mass}}{\text{molecular mass}} = \frac{50 \text{ g}}{32 \text{ g/mol}} = 1.5625 \text{ mol} \]

### Step 2: Determine the initial temperature (T₁)
At Normal Temperature and Pressure (NTP), the temperature is given as:

\[ T_1 = 273 \text{ K} \]

### Step 3: Use the adiabatic relation to find the final temperature (T₂)
The adiabatic relation between pressure and temperature is given by:

\[ \frac{P_1}{P_2} = \left(\frac{T_1}{T_2}\right)^{\frac{\gamma}{\gamma - 1}} \]

Where:
- \(P_1 = 1 \text{ atm}\) (initial pressure at NTP)
- \(P_2 = 5 \text{ atm}\) (final pressure)
- \(\gamma = 1.4\)

Substituting the values, we have:

\[ \frac{1}{5} = \left(\frac{273}{T_2}\right)^{\frac{1.4}{0.4}} \]

This simplifies to:

\[ \frac{1}{5} = \left(\frac{273}{T_2}\right)^{3.5} \]

Taking the reciprocal and raising both sides to the power of \(\frac{1}{3.5}\):

\[ \frac{T_2}{273} = 5^{\frac{1}{3.5}} \]

Calculating \(5^{\frac{1}{3.5}} \approx 1.903\):

\[ T_2 = 273 \times 1.903 \approx 519.2 \text{ K} \]

### Step 4: Calculate the work done (W)
The work done on the gas during adiabatic compression is given by:

\[ W = nR \frac{(T_2 - T_1)}{\gamma - 1} \]

Where:
- \(R = 8.31 \text{ J/(mol K)}\)
- \(T_1 = 273 \text{ K}\)
- \(T_2 \approx 519.2 \text{ K}\)

Substituting the values:

\[ W = 1.5625 \times 8.31 \times \frac{(519.2 - 273)}{0.4} \]

Calculating the temperature difference:

\[ T_2 - T_1 = 519.2 - 273 = 246.2 \text{ K} \]

Now substituting:

\[ W = 1.5625 \times 8.31 \times \frac{246.2}{0.4} \]

Calculating further:

\[ W = 1.5625 \times 8.31 \times 615.5 \approx 1.5625 \times 5112.83 \approx 7999.99 \text{ J} \]

Since work done on the gas is considered negative during compression:

\[ W \approx -8000 \text{ J} \]

### Final Result
The work done on the gas during adiabatic compression is approximately \(-8000 \text{ J}\).

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