Question 170: Thermodynamics & KTG

Question

A total of $$48 \mathrm{~J}$$ heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by $$2^{\circ} \mathrm{C}$$. The work done by the gas is:
Given, $$\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$$.

Accepted Answer

To solve this problem, we can use the first law of thermodynamics, which states:
$$\Delta Q = \Delta U + W$$
where:
$$\Delta Q$$ is the heat transferred to the system, in this case, $$48 \mathrm{~J}$$.
$$\Delta U$$ is the change in internal energy of the system.
$$W$$ is the work done by the system.
For one mole of an ideal gas, the change in internal energy can be calculated using the equation:
$$\Delta U = nC_{v}\Delta T$$
where:
$$n$$ is the number of moles, which is $$1$$ in this case.
$$C_{v}$$ is the molar specific heat capacity at constant volume.
$$\Delta T$$ is the change in temperature, in Kelvins.
Given that we are dealing with helium, a monatomic ideal gas, we know that:
$$C_{v} = \frac{3}{2}R$$
Substituting the given values, we get:
$$\Delta U = (1) \frac{3}{2}(8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1})(2^{\circ} \mathrm{C})$$
Since a change in temperature of $$1^{\circ} \mathrm{C}$$ is equivalent to a change in temperature of $$1 \mathrm{~K}$$, we can directly substitute $$2^{\circ} \mathrm{C}$$ as a $$2 \mathrm{~K}$$ change without needing to convert Celsius to Kelvin as both scales have the same magnitude of degree.
Therefore,
$$\Delta U = 1 	imes \frac{3}{2} 	imes 8.3 	imes 2 = 24.9 \mathrm{~J}$$
Now, substituting $$\Delta U$$ and $$\Delta Q$$ into the first law of thermodynamics:
$$48 = 24.9 + W$$
Solving for $$W$$ we find:
$$W = 48 - 24.9 = 23.1 \mathrm{~J}$$
Thus, the work done by the gas is $$23.1 \mathrm{~J}$$, which corresponds to Option A.

Answer

23.1 J

Answer

48 J

Answer

24.9 J

Answer

72.9 J

Thermodynamics & KTG

A total of $48 \mathrm{~J} heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by 2^{\circ} \mathrm{C}. The work done by the gas is: Given, \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.

Choose the correct answer: