Question

When the temperature of a gas, contained in a closed vessel is increased by $5C$, it’s pressure increases by $1$ %. The original temperature of the gas was approximately?
A) $500$
B) $273$
C) $227$
D) $50$

Answer 1

Hint: Ideal gas law states that the state of an amount of gas is evaluated by its pressure, volume, and temperature of gas
Ideal gas equation is \[pV{\text{ }} = {\text{ }}nRT\]
Where, p is pressure, V is volume of gas, n is amount of substance of gas, R is called as gas constant
T is temperature of amount of gas
In SI units, Pressure is measured in Pascals, amount of substance of gas is measured in moles, Volume is measured in cubic metres, and Temperature is measured in kelvin

Formula used: $\dfrac{{{P_1}}}{{{T_1}}} = \dfrac{{{P_2}}}{{{T_2}}}$

Complete step-by-step answer:
It is constant volume process,
Hence apply the formula,
$\dfrac{{{P_1}}}{{{T_1}}} = \dfrac{{{P_2}}}{{{T_2}}}$
We put, ${P_1} = 1$ for pressure
As pressure increases by \[1\% \]
${P_2} = P + 1\% $ of \[P\]= $1.01{P_1}$
${T_1}$ for temperature
${T_2}$ = $\left( {{T_1} + 5} \right)$
Substitute the values in formula,
$\Rightarrow$\[\;\dfrac{1}{{1.01}}\]${P_1}$ = $\dfrac{{{T_1}}}{{\left( {{T_1} + 5} \right)}}$
To find ${T_1}$ we have to bring all the values from RHS TO LHS
$\Rightarrow$\[\;\dfrac{1}{{1.01}}\]$\left( {{T_1} + 5} \right) = {T_1}$
By dividing we get \[\;\dfrac{1}{{1.01}}\]
$\Rightarrow$${T_1}$ = $0.990$ \[(T + 5)\;\]
Multiplying the values and we get,
$\Rightarrow$${T_1}$ = $0.990{T_1}$ + $4.95$
$\Rightarrow$${T_1}$= $500K$
Subtract $273$ to get Celsius and the ${T_1}$=$227^\circ C$
By applying ideal gas laws we got the initial temperature value as ${T_1}$=$227^\circ C$ by converting the kelvin value to Celsius

Hence the correct option is (C).

Note: The ideal gas law is called the general gas equation.
This state of equation is a hypothetical ideal gas.
The number of moles, n, is fixed because the temperature of gas is in a closed vessel. The volume, V is also fixed.