Question 720: Thermodynamics & KTG

Question

A gas is contained in closed vessel. The initial temperature of the gas is $100^{\circ} \mathrm{C}$. If the pressure of the gas is increased by $4 \%$, the increase in the temperature of the gas is

Accepted Answer

To find the increase in the temperature of the gas, we start with the initial conditions and apply the principles of ideal gas law. The initial pressure is given as $ P_1 $, and the pressure increase is 4%, so the new pressure is:
$  P_2 = P_1 + 0.04 P_1 = 1.04 P_1  $
Using the ideal gas equation $ PV = nRT $, we relate the initial and final states as follows, assuming the volume and the amount of gas remain constant:
$  \frac{T_1}{T_2} = \frac{P_1}{P_2}  $
The initial temperature $ T_1 $ is given as $ 100^{\circ} \mathrm{C} $, which converts to Kelvin as $ 373 \, \mathrm{K} $.
Supposing the final temperature is $ T_2 = T_1 + \Delta T $:
$  \frac{373}{373 + \Delta T} = \frac{1}{1.04}  $
By cross-multiplying, we solve for $\Delta T$:
$  373(1.04) = 373 + \Delta T  $
$  387.92 = 373 + \Delta T  $
$  \Delta T = 387.92 - 373 = 14.92 \, \mathrm{K}  $
Since temperature changes in Celsius and Kelvin are equivalent:
$  \Delta T = 14.92^{\circ} \mathrm{C}  $
Thus, the increase in the temperature of the gas is $ 14.92^{\circ} \mathrm{C} $.

Answer

$2^{\circ} \mathrm{C}$

Answer

$3^{\circ} \mathrm{C}$

Answer

$4^{\circ} \mathrm{C}$

Answer

$5^{\circ} \mathrm{C}$

Thermodynamics & KTG

A gas is contained in closed vessel. The initial temperature of the gas is 100^{\circ} \mathrm{C}. If the pressure of the gas is increased by 4 \%, the increase in the temperature of the gas is

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