What is the volume at `N.T.P.` of a gas that occupies `43.0 mL` at `- 3^(@)C` and `0.98 "bar"` ?

Explanation

To find the volume of a gas at Normal Temperature and Pressure (NTP) given its initial conditions, we can use the combined gas law, which relates pressure, volume, and temperature. Here’s how to solve the problem step by step:

### Step 1: Identify the given values
- Initial volume (V1) = 43.0 mL
- Initial temperature (T1) = -3°C
- Initial pressure (P1) = 0.98 bar
- Normal temperature (T2) = 273 K (NTP)
- Normal pressure (P2) = 1.013 bar (NTP)

### Step 2: Convert the initial temperature to Kelvin
To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273.15 \]
So,
\[ T1 = -3 + 273.15 = 270.15 \, K \]

### Step 3: Use the combined gas law
The combined gas law is given by:
\[ \frac{P1 \cdot V1}{T1} = \frac{P2 \cdot V2}{T2} \]
Rearranging this equation to solve for V2 gives:
\[ V2 = \frac{P1 \cdot V1 \cdot T2}{P2 \cdot T1} \]

### Step 4: Substitute the known values into the equation
Now we can substitute the values into the equation:
- P1 = 0.98 bar
- V1 = 43.0 mL
- T2 = 273 K
- P2 = 1.013 bar
- T1 = 270.15 K

\[ V2 = \frac{0.98 \, \text{bar} \cdot 43.0 \, \text{mL} \cdot 273 \, K}{1.013 \, \text{bar} \cdot 270.15 \, K} \]

### Step 5: Calculate V2
Now, we can perform the calculation:
1. Calculate the numerator:
\[ 0.98 \cdot 43.0 \cdot 273 \approx 1.141 \, \text{(approximately)} \]
2. Calculate the denominator:
\[ 1.013 \cdot 270.15 \approx 273.5 \, \text{(approximately)} \]
3. Divide the numerator by the denominator:
\[ V2 \approx \frac{1.141}{273.5} \approx 42.06 \, \text{mL} \]

### Final Answer
The volume of the gas at NTP is approximately **42.06 mL**.

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