Pressure of an ideal gas, contained in a closed vessel, is increased by 0.4% when heated by $ 1^\circ \text{C} $. Its initial temperature must be :

Question

Given $C_p = a + bT$
$a = 19.5$, $b = 0.042$, $n = 1$ mole
Temperature changes from $27^\circ C$ to $327^\circ C$.
Find $\Delta H$.

Answer 1

This problem requires the application of the ideal gas law, specifically the pressure-temperature relationship. Given that a 0.4% increase in pressure accompanies a $ 1^\circ \text{C} $ temperature increase, the objective is to determine the initial temperature.

The ideal gas law is stated as:

$PV = nRT$

In this equation, $P$ denotes pressure, $V$ represents volume, $n$ signifies the number of moles, $R$ is the gas constant, and $T$ is the temperature in Kelvin.

As the vessel is sealed, both volume $V$ and number of moles $n$ remain constant. Consequently, the direct relationship between pressure and temperature is:

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

Where:

$P_1$ is the initial pressure
$T_1$ is the initial temperature
$P_2$ is the final pressure
$T_2$ is the final temperature

The problem specifies a 0.4% pressure increase, leading to:

$P_2 = P_1 \times 1.004$

A $1^\circ \text{C}$ temperature increase means the final temperature $T_2$ is:

$T_2 = T_1 + 1$

Substituting these into the derived relation:

$\frac{P_1}{T_1} = \frac{P_1 \times 1.004}{T_1 + 1}$

Upon cancelling $P_1$ from both sides:

$\frac{1}{T_1} = \frac{1.004}{T_1 + 1}$

Cross-multiplication yields:

$T_1 + 1 = T_1 \times 1.004$

Rearranging the equation to solve for $T_1$:

$1 = 0.004 \times T_1$

Therefore:

$T_1 = \frac{1}{0.004}$

$T_1 = 250 \, \text{K}$

The initial temperature of the gas was $ 250 \, \text{K} $. The correct answer is: $ 250 \, \text{K} $

Pressure of an ideal gas, contained in a closed vessel, is increased by 0.4% when heated by \( 1^\circ \text{C} \). Its initial temperature must be :

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The Correct Option is C

Solution and Explanation

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