Physics Thermodynamics

$\mathrm{T}\mathrm{i}=-{50}^{?}\mathrm{C}$

$\begin{array}{rr}& =223\mathrm{K}\\ & {\mathrm{V}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\propto \sqrt[]{\mathrm{T}}\end{array}$

As ${\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}$  increased by 3 times

So ${\left({\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\right)}_{\mathrm{f}}=4{\left({\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\right)}_{\text{initial}}$

${\mathrm{T}}_{\mathrm{f}}=16{\mathrm{T}}_{\mathrm{i}}$

$=16*223$

$=3568\mathrm{K}$ ${\mathrm{T}}_{\mathrm{f}}=(3568-273{)}^{?}\mathrm{C}$

$={3295}^{?}\mathrm{C}$

15. (c) Fundamental harmonic frequency open pipe

$=\frac{\mathrm{V}}{2\mathrm{L}}={\mathrm{v}}_1$

(say)

Fundamental harmonic frequency of closed pipe $=\frac{\mathrm{V}}{4\mathrm{L}}$

$={\mathrm{v}}_2$ (say)

$\Rightarrow \frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\frac{\frac{\mathrm{V}}{2\mathrm{L}}}{\frac{\mathrm{V}}{4\mathrm{L}}}\mathrm{}2:1$

Kindly go through the answers

S? * S?

PV? = C ⇒ V? (dP/dV) + P (γV^ (γ-1) = 0 ⇒ dP/dV = -γ (P/V) ⇒ dP/P = -γ (dV/V)

Pressure decreases with an increase in volume in both isothermal and adiabatic processes. In an adiabatic process, as volume decreases, pressure increases with the increase in temperature.

In an LCR series AC circuit, the phase difference φ between current and voltage is: φ = tan? ¹ (X? - X? ) / R)

If each vibrational mode contributes two degrees of freedom and the total degrees of freedom f = 3 + 3 + 4 = 10, then:
β = 1 + 2/f = 1 + 2/10 = 1.2

For adiabatic process CD,
W = (P? V? - P? V? ) / (γ - 1) = (100 * 4 - 200 * 3) / (1.4 - 1) = -500J

U = U? + U? = (n? /N_A) (F? R/2)T? + (n? /N_A) (F? R/2)T?

For the mixture: U = (n? +n? )/N_A * (FR/2)T

F = (n? F? + n? F? ) / (n? + n? )

Equating the expressions for U and solving for T gives:
T = (n? F? T? + n? F? T? ) / (n? F? + n? F? )

η = 1 - T_C / T_H

η = 1 - 400/800 = 1 - ½ = ½

η = W/Q_H ⇒ ½ = W/Q_H ⇒ Q_H = 2W = 2 * 1200 = 2400 J

f = f_Translational + f_Rotational + f_Vibrational

f = 3 + 3 + 48 = 54 (Since each vibrational mode has two degrees of freedom)

γ = 1 + 2/f = 1 + 2/54 = 1 + 1/27 = 28/27 ≈ 1.03

1/4 m (210)² = m (0.03) x (4.2) x 1000 x ΔT ; Q = mSΔt
ΔT = (210) (210)/ (4) (4.2) (0.03) (1000) = 87.5°C