the first law of thermodynamics

Question 1

An ideal gas of mass $$0.35 \mathrm{~kg}$$ is heated at a constant pressure of $$2.0 \times 10^5 \mathrm{~Pa}$$ so that its internal energy increases by $$7600 \mathrm{~J}$$．During this process，the volume of the gas increases from $$0.038 \mathrm{~m}^3$$ to $$0.063 \mathrm{~m}^3$$ and the temperature increases by $$56^{\circ} \mathrm{C}$$．（i）Show that the magnitude of the work done on the gas is $$5000 \mathrm{~J}$$．[1] （ii）Explain whether the work done on the gas is positive or negative．[2] （iii）Determine the magnitude of the thermal energy $$q$$ transferred to the gas．[2] （iv）Calculate the specific heat capacity of the gas for this process．Give a unit with your answer．[2]

Answer

Answer:

略

Explanation:

Step 1. Understanding the Work Done on the Gas

To find the work done on the gas, we use the formula for work done at constant pressure:

$$ \text{Work done} = p \Delta V $$

where:
- $$ p $$ is the pressure,
- $$ \Delta V $$ is the change in volume.

Given:
- $$ p = 2.0 \times 10^5 \, \text{Pa} $$,
- Initial volume $$ V_i = 0.038 \, \text{m}^3 $$,
- Final volume $$ V_f = 0.063 \, \text{m}^3 $$.

Calculate the change in volume:

$$ \Delta V = V_f - V_i = 0.063 - 0.038 = 0.025 \, \text{m}^3 $$

Now, calculate the work done:

$$ \text{Work done} = 2.0 \times 10^5 \times 0.025 = 5000 \, \text{J} $$

Step 2. Determining the Sign of the Work Done

Since the gas is expanding against an external pressure, the work is done by the gas on the surroundings.

Therefore, the work done on the gas is negative.

Step 3. Calculating the Thermal Energy Transferred

The first law of thermodynamics relates the change in internal energy ($$ \Delta U $$), the heat added to the system ($$ q $$), and the work done by the system ($$ W $$):

$$ \Delta U = q + W $$

Given:
- $$ \Delta U = 7600 \, \text{J} $$,
- Work done by the gas $$ W = -5000 \, \text{J} $$ (since work done on the gas is negative).

Rearrange the formula to solve for $$ q $$:

$$ q = \Delta U - W = 7600 - (-5000) = 7600 + 5000 = 12600 \, \text{J} $$

Step 4. Calculating the Specific Heat Capacity

The specific heat capacity ($$ c $$) is calculated using the formula:

$$ c = \frac{q}{m \Delta T} $$

where:
- $$ q = 12600 \, \text{J} $$,
- $$ m = 0.35 \, \text{kg} $$,
- $$ \Delta T = 56 \, \text{K} $$ (since the temperature change in Celsius is equivalent to Kelvin).

Substitute the values into the formula:

$$ c = \frac{12600}{0.35 \times 56} $$

Calculate:

$$ c = \frac{12600}{19.6} = 642.86 \, \text{J} \, \text{kg}^{-1} \, \text{K}^{-1} $$

Rounding to three significant figures, the specific heat capacity is:

$$ c = 640 \, \text{J} \, \text{kg}^{-1} \, \text{K}^{-1} $$

This completes the problem-solving process, confirming the calculations and understanding of the thermodynamic principles involved.

Question 2

The gas in（b）is now heated at constant volume rather than at constant pressure． The increase in internal energy of the gas is the same as in（b）． Use the first law of thermodynamics to explain whether the specific heat capacity of the gas for this process is less than，the same as，or greater than the answer in（b）（iv）．[3]

Answer

Answer:

略

Explanation: