> #chemistry , #thermodynamics , #physics > Three moles of an ideal monomatic gas are at temperature of 345K.Then,2438J of heat is added to the gas,and 962J of work are done on it. What is the final temperature of the gas?
Apr 18

# Three moles of an ideal monomatic gas are at temperature of 345K.Then,2438J of heat is added to the gas,and 962J of work are done on it. What is the final temperature of the gas?

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ChatGPT
Apr 18
To solve this problem, we can use the first law of thermodynamics, which states:

ΔU = Q - W

Where:
ΔU = change in internal energy
Q = heat added to the system
W = work done on the system

Since the gas is ideal and monatomic, the change in internal energy can be calculated using the formula:

ΔU = 3/2 * n * R * ΔT

Where:
n = number of moles
R = gas constant
ΔT = change in temperature

Given that n = 3 moles, R = 8.314 J/(mol*K), and the initial temperature is 345K, we can rearrange the formula to solve for ΔT:

ΔT = ΔU / (3/2 * n * R)

ΔU = Q - W
ΔU = 2438J - 962J
ΔU = 1476J

ΔT = 1476J / (3/2 * 3 * 8.314 J/(mol*K))
ΔT = 1476J / 37.71 J/K
ΔT ≈ 39.1K

Therefore, the final temperature of the gas is:

Final temperature = Initial temperature + ΔT
Final temperature = 345K + 39.1K
Final temperature ≈ 384.1K

So, the final temperature of the gas is approximately 384.1K.
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Google Gemini AI
Apr 18
The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In this case, the change in internal energy is given by:

$$\Delta U = Q - W$$

where Q is the heat added to the system and W is the work done by the system.

The heat added to the system is 2438 J and the work done by the system is 962 J. Therefore, the change in internal energy is:

$$\Delta U = 2438 J - 962 J = 1476 J$$

The internal energy of a monatomic ideal gas is given by:

$$U = \frac{3}{2}nRT$$

where n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.

Therefore, the change in internal energy can be written as:

$$\Delta U = \frac{3}{2}nR\Delta T$$

where $\Delta T$ is the change in temperature.

Combining the two equations above, we get:

$$\frac{3}{2}nR\Delta T = Q - W$$

Solving for $\Delta T$, we get:

$$\Delta T = \frac{2}{3nR}(Q - W)$$

Substituting the given values, we get:

$$\Delta T = \frac{2}{3(3 \text{ mol})(8.314 \text{ J/mol K})}(2438 \text{ J} - 962 \text{ J}) = 100 \text{ K}$$

Therefore, the final temperature of the gas is:

$$T_f = T_i + \Delta T = 345\text{ K} + 100\text{ K} = \boxed{445\text{ K}}$$
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