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8. A mass m_{g} = 50g of an ideal gas is held at a temperature of t_{1} = 0 ^ 0 * C in a container of constant volume. The gas absorbs a quantity of heat Delta*Q = 1.25 * 10 ^ 5 * J and as a result its pressure increases to three times its initial value. What is the final temperature t2 of the gas? What is its specific heat at constant volume C, (in J.kg¹)?
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8. A mass m_{g} = 50g of an ideal gas is held at a temperature of t_{1...
Understanding the Problem
To determine the final temperature (t2) of the gas and its specific heat at constant volume (C), we will utilize the ideal gas law and the relationship between heat, temperature change, and specific heat.
Given Data
- Mass of gas (m_g) = 50 g = 0.05 kg
- Initial temperature (t1) = 0 °C = 273 K
- Heat absorbed (Delta Q) = 1.25 x 10^5 J
- Final pressure (P2) = 3 x Initial pressure (P1)
Using the Ideal Gas Law
In a constant volume process, the relationship between heat added and temperature change is given by:
Delta Q = m * C * Delta T
Where:
- Delta T = t2 - t1
Since pressure increases threefold, we can relate the temperatures using the ideal gas law:
P1/T1 = P2/T2
Substituting P2 = 3 * P1:
1/T1 = 3/T2
This gives us:
T2 = 3 * T1
Calculating Final Temperature (t2)
Substituting the initial temperature:
T2 = 3 * 273 K = 819 K
Now, converting back to Celsius:
t2 = 819 K - 273 = 546 °C
Finding Specific Heat (C)
Substituting known values into the heat equation:
1.25 x 10^5 J = 0.05 kg * C * (819 K - 273 K)
1.25 x 10^5 J = 0.05 kg * C * 546 K
Solving for C:
C = (1.25 x 10^5 J) / (0.05 kg * 546 K) = 4,578.5 J/kg·K
Final Results
- Final Temperature (t2) = 546 °C
- Specific Heat (C) = 4,578.5 J/kg·K
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8. A mass m_{g} = 50g of an ideal gas is held at a temperature of t_{1} = 0 ^ 0 * C in a container of constant volume. The gas absorbs a quantity of heat Delta*Q = 1.25 * 10 ^ 5 * J and as a result its pressure increases to three times its initial value. What is the final temperature t2 of the gas? What is its specific heat at constant volume C, (in J.kg¹)?
Question Description
8. A mass m_{g} = 50g of an ideal gas is held at a temperature of t_{1} = 0 ^ 0 * C in a container of constant volume. The gas absorbs a quantity of heat Delta*Q = 1.25 * 10 ^ 5 * J and as a result its pressure increases to three times its initial value. What is the final temperature t2 of the gas? What is its specific heat at constant volume C, (in J.kg¹)? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about 8. A mass m_{g} = 50g of an ideal gas is held at a temperature of t_{1} = 0 ^ 0 * C in a container of constant volume. The gas absorbs a quantity of heat Delta*Q = 1.25 * 10 ^ 5 * J and as a result its pressure increases to three times its initial value. What is the final temperature t2 of the gas? What is its specific heat at constant volume C, (in J.kg¹)? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 8. A mass m_{g} = 50g of an ideal gas is held at a temperature of t_{1} = 0 ^ 0 * C in a container of constant volume. The gas absorbs a quantity of heat Delta*Q = 1.25 * 10 ^ 5 * J and as a result its pressure increases to three times its initial value. What is the final temperature t2 of the gas? What is its specific heat at constant volume C, (in J.kg¹)?.
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