If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by 1°C, the initial temperature must be

The correct option is 250 K.

Step-by-Step Explanation:

1. Identify the physical process:
The gas is contained in a closed vessel, which means its volume ($V$) remains constant. Since the volume and the amount of gas are constant, we can apply Gay-Lussac's Law (also known as the Pressure Law).
According to Gay-Lussac's Law, the pressure ($P$) of a given mass of gas is directly proportional to its absolute temperature ($T$ in Kelvin):

$P\propto T$

This can be written as:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where:
- $P_1$ is the initial pressure,
- $T_1$ is the initial absolute temperature (in Kelvin),
- $P_2$ is the final pressure,
- $T_2$ is the final absolute temperature (in Kelvin).

2. Define the given variables:
Let the initial pressure be $P_1=P$ and the initial temperature be $T_1=T$.
The gas is heated by $1°C$. A change in temperature of $1°C$ is equal to a change of $1K$. Thus, the final temperature is:

$T_2=T+1$

The pressure increases by $0.4\%$. Therefore, the final pressure $P_2$ is:

$P_2=P+0.4\%ofP=P+\frac{0.4}{100}P=P(1+0.004)=1.004P$

3. Substitute the values into the formula and solve for $T$:

$\frac{P}{T}=\frac{1.004P}{T+1}$

Since $P$ is non-zero, we can cancel $P$ from both sides:

$\frac{1}{T}=\frac{1.004}{T+1}$

Cross-multiplying to solve for $T$:

$T+1=1.004T$

Subtract $T$ from both sides:

$1=1.004T-T$

$1=0.004T$

$T=\frac{1}{0.004}$

$T=\frac{1000}{4}=250K$

Thus, the initial temperature of the gas must be 250 K.