Question 620: Thermodynamics & KTG

Question

When the pressure of the gas contained in a closed vessel is increased by $2.3 \%$, the temperature of the gas increases by 4 K . The initial temperature of the gas is

Accepted Answer

We are given:
A closed vessel containing some gas.
Pressure increases by 2.3%.
Resulting temperature increases by 4 K.
We need the initial temperature.
Step 1: Use Ideal Gas Law
For a fixed volume and amount of gas:
$ \frac{P}{T} = 	ext{constant} $
So,
$ \frac{P_2}{P_1} = \frac{T_2}{T_1}. $
Step 2: Express the pressure change
Pressure increases by 2.3%, i.e.
$ \frac{P_2}{P_1} = 1.023. $
Step 3: Express the temperature change
$T_2 = T_1 + 4$.
So,
$ \frac{P_2}{P_1} = \frac{T_1 + 4}{T_1}. $
Step 4: Write the equation
$ 1.023 = \frac{T_1 + 4}{T_1} = 1 + \frac{4}{T_1}. $
Step 5: Solve for $T_1$
$ 1.023 - 1 = \frac{4}{T_1}, $
$ 0.023 = \frac{4}{T_1}. $
$ T_1 = \frac{4}{0.023}. $
$ T_1 \approx 173.9 \, 	ext{K}. $
Step 6: Match to options
Closest option is
160 K (Option C)
.
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Final Answer: 160 K (Option C)

Answer

80 K

Answer

150 K

Answer

160 K

Answer

320 K

Thermodynamics & KTG

When the pressure of the gas contained in a closed vessel is increased by 2.3 \%, the temperature of the gas increases by 4 K . The initial temperature of the gas is

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