Question

1. At what temperature the root mean square velocity will be half of that standard pressure
and temperature, the pressure being kept constant.
2. For an ideal gas γ=1.4, calculate the values of molar specific heats of the gas. (R=8.31
J/Kmol)
3. How much heat is required to raise the temperature by 40 of 14g nitrogen gas at constant
pressure?[Molar mass of nitrogen=28 g, R=8.31J/Kmol for diatomic gas =5/2R]
4. In Young’s experiment separation between two slits is 3.5×10 -4 m and the distance of the
screen from the plane of the slit is .73m. What is the distance from the central bright
point to the first bright point in the screen? [λ=550×10 -10 m]
5. The phase difference between two points in a wave is /2. What is the path difference
between those two points?

If any question need explanation then those can be ignored. only the mathematical questions needs to be solved. thank you

Answer 1

Answer on Question #83474 - physics - thermodynamics

1. At what temperature the root mean square velocity will be half of that standard pressure and temperature, the pressure being kept constant.

Solution.

Calculate the root mean square velocity by relation as follows.

$$V_{rms} = \sqrt{\frac{3RT}{M}}$$

Here, $R$ is the gas constant, $T$ is the standard temperature and $M$ is the mass of mole.

Here, $V_{rms}$ is directly proportional to the square root of temperature.

$$V_{rms} \propto \sqrt{T}$$

Substitute 273 K for T.

Calculate the temperature at half root mean square velocity by the relation as follows.

$$V_{rms}' \propto \sqrt{T'}$$

Here, $V_{rms}'$ half mean square velocity.

By the proportionality relation

$$\frac{V_{rms}}{V_{rms}'} = \sqrt{\frac{273\,K}{T'}} = 2.25\,K$$ $$2 = \sqrt{\frac{273\,K}{T'}} = 2.25\,K$$ $$T' = 68.25\,K$$

Answer: 68.25 K

2. For an ideal gas = 1.4, calculate the values of molar specific heats of the gas. (R=8.31 J/mol K)

Solution.

Calculate the molar specific heats by the relation as follows.

$$\gamma = \frac {C _ {P}}{C _ {V}}$$ $$C _ {P} = \gamma C _ {V}$$

Here, $C_P$ is the molar specific heat at the constant pressure and $C_V$ is the molar specific heat at constant volume.

Calculate the values of molar specific heats by the relation as follows.

$$C _ {P} - C _ {V} = R$$ $$\gamma C _ {V} - C _ {V} = R$$

Substitute $\gamma = 1.4$ and R=8.31 J/mol K.

$$1.4 C _ {V} - C _ {V} = 8.31$$ $$C _ {V} = \frac {8.31 \ \mathrm{J/molK}}{0.4}$$ $$C _ {V} = 20.775 \ \mathrm{J/molK}$$

Calculate $C_P$ by the relation

$$C _ {P} = \gamma C _ {V}$$ $$C _ {P} = (1.4) \left(20.775 \ \frac{\mathrm{J}}{\mathrm{molK}}\right)$$ $$C _ {P} = 29.085 \ \mathrm{J/molK}$$

Answer.

$$C _ {V} = 20.775 \ \mathrm{J/molK}$$ $$C _ {P} = 2 9. 0 8 5 \frac {J}{m o l} K$$

3. How much heat is required to raise the temperature by 40 of 14g nitrogen gas at constant pressure? [Molar mass of nitrogen=28 g, R=8.31J/Kmol for diatomic gas =5/2R]

Solution.

Calculate the heat required by the relation as follows.

$$Q = n C _ {V} \Delta T$$

Here, $n$ is the mole of gas, $C_V$ is the molar specific heat of gas and $\Delta T$ is the temperature difference.

$$Q = (14/28)(5R/2)(40)$$ $$Q = (0.5)\left(2.5 \times 8.31 \frac{J}{mol} K\right)(40 K)$$ $$Q = 415.7 J$$

Answer. 415.7 J

4. In Young's experiment separation between two slits is $3.5 \times 10^{-4} \, \text{m}$ and the distance of the

screen from the plane of the slit is .73m. What is the distance from the central bright

point to the first bright point in the screen? $[\lambda = 550 \times 10^{-10} \, \text{m}]$

Solution.

Calculate the distance of first bright point from the central bright point by the relation as follows.

$$x = \left(\frac {D}{d}\right) n \lambda$$

Substitute $0.73\mathrm{m}$ for D and $3.5\times 10^{-4}\mathrm{m}$ for d and $550\times 10^{-10}\mathrm{m}$ for $\lambda$ and 1 for n.

$$\begin{array}{l} x = \left(\frac {0 . 7 3 \mathrm {m}}{3 . 5 \times 1 0 ^ {- 4} \mathrm {m}}\right) (1) (5 5 0 \times 1 0 ^ {- 1 0} \mathrm {m}) \\ x = 1 1 4. 7 1 \times 1 0 ^ {- 6} m \\ \end{array}$$

Answer. $114.71 \times 10^{-6} m$

5. The phase difference between two points in a wave is $\pi/2$. What is the path difference between those two points?

Solution.

Calculate the path difference by the relation as follows.

$$\Delta X = \frac{\lambda \cdot \Delta\phi}{2\pi}$$

Here, $\lambda$ is the wavelength, $\Delta\phi$ is the phase difference.

Substitute $\frac{\pi}{2}$ for phase difference.

$$\Delta X = \frac{\lambda \cdot \left(\frac{\pi}{2}\right)}{2\pi}$$ $$\Delta X = \frac{\lambda}{4}$$

Answer. $\frac{\lambda}{4}$

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