Problem 91
Question
When a gas filled in a closed vessel is heated through \(1^{\circ} \mathrm{C}\), its pressure increases by \(0.4 \%\). The initial temperature of the gas was (a) \(250 \mathrm{~K}\) (b) \(2500 \mathrm{~K}\) (c) \(250^{\circ} \mathrm{C}\) (d) \(25^{\circ} \mathrm{C}\)
Step-by-Step Solution
Verified Answer
The initial temperature of the gas was 250 K.
1Step 1: Understand the problem
The problem is about the relationship between temperature and pressure in a gas that's kept in a closed container. When heated by 1°C, the gas pressure increases by 0.4%. This is a due to one of the gas laws.
2Step 2: Apply Gay-Lussac's Law
Gay-Lussac's Law states that pressure is directly proportional to temperature when volume is kept constant, represented as \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). Here, \(P_2 = P_1(1 + 0.004)\) since pressure increases by 0.4%. We need to find \(T_1\), the initial temperature.
3Step 3: Set up the equation
With Gay-Lussac's formula, substitute: \( \frac{P}{T} = \frac{P(1+0.004)}{T+1} \). Simplify to find the initial temperature.
4Step 4: Solve for Initial Temperature \(T\)
Rearranging the equation gives \( T(1 + 0.004) = T+1 \). Solving gives: \( 0.004T = 1 \) thus, \( T = \frac{1}{0.004} = 250 \).
5Step 5: Interpret the result
The initial temperature was found to be 250 K based on the calculations. This corresponds to option (a).
Key Concepts
Temperature-Pressure RelationshipGas LawsInitial Temperature Calculation
Temperature-Pressure Relationship
Understanding the temperature-pressure relationship in gases is crucial for grasping Gay-Lussac's Law. When a gas is contained within a closed vessel, its pressure tends to change if the temperature changes.
This occurs because molecules move more quickly and collide with the walls of the vessel more frequently, resulting in higher pressure. In this exercise, we learned that a 1°C rise in temperature causes a 0.4% increase in pressure. This indicates how temperature can substantially impact the pressure of a gas. Some key points about this relationship include:
This occurs because molecules move more quickly and collide with the walls of the vessel more frequently, resulting in higher pressure. In this exercise, we learned that a 1°C rise in temperature causes a 0.4% increase in pressure. This indicates how temperature can substantially impact the pressure of a gas. Some key points about this relationship include:
- Temperature and pressure are directly proportional.
- As the temperature rises, the kinetic energy of gas molecules increases.
- Faster moving molecules lead to increased collisions within the vessel, raising the pressure.
Gas Laws
Gas laws are fundamental concepts that describe how gases behave under various conditions. One of these essential laws is Gay-Lussac's Law.Gay-Lussac's Law specifically examines the relationship between temperature and pressure when the volume is kept constant. The equation representing this law is:\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]It states that the ratio of the initial pressure to initial temperature is equal to the ratio of the final pressure to final temperature.
This relationship allows us to predict how pressure changes with temperature in a controlled environment.
**Key takeaways from gas laws include:**
This relationship allows us to predict how pressure changes with temperature in a controlled environment.
**Key takeaways from gas laws include:**
- They assume ideal conditions where gases behave predictably.
- They help calculate changes in gas properties under different scenarios.
- They inform practical applications, such as cooking or inflating tires.
Initial Temperature Calculation
Calculating the initial temperature of a gas using Gay-Lussac's Law can be a straightforward process with practice.In this scenario, we used the given increase in pressure to determine the starting temperature. By applying the law's formula:\[ \frac{P}{T} = \frac{P(1+0.004)}{T+1} \]We simplified to derive:\[ T(1 + 0.004) = T+1 \]To solve for the initial temperature \( T \), follow these steps:
- Rearrange the formula to emphasize \( T \).
- Calculate the change: \( 0.004T = 1 \).
- Find \( T \) by \( T = \frac{1}{0.004} = 250 \).
Other exercises in this chapter
Problem 89
1 mole of \(\mathrm{H}_{2}\) gas is contained in a box of volume \(V=1.00 \mathrm{~m}^{3}\) at \(T=300 \mathrm{~K}\). The gas is heated to a temperature of \(T=
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