The ratio of the specific heat capacities of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given,the ratio of specific heat capacities $\gamma = 1.5 = 3/2$.For an adiabatic process,$P_i V_i^\gamma = P_f V_f^\gamma$. Given $V_f = 2V_i$ an

Vedclass · Accepted Answer

$\sqrt{2}: 1$

Vedclass · Answer

(D) Given,the ratio of specific heat capacities $\gamma = 1.5 = 3/2$.For an adiabatic process,$P_i V_i^\gamma = P_f V_f^\gamma$. Given $V_f = 2V_i$ and $P_f = P_1$,we have $P_{i, 	ext{ad}} V_i^{1.5} = P_1 (2V_i)^{1.5}$.Thus,$P_{i, 	ext{ad}} = P_1 (2)^{1.5} = P_1 (2\sqrt{2})$.For an isothermal process,$P_i V_i = P_f V_f$. Given $V_f = 2V_i$ and $P_f = P_2$,we have $P_{i, 	ext{iso}} V_i = P_2 (2V_i)$.Thus,$P_{i, 	ext{iso}} = 2P_2$.Given $P_1 = P_2$,the ratio of initial pressures is $\frac{P_{i, 	ext{ad}}}{P_{i, 	ext{iso}}} = \frac{P_1 (2)^{1.5}}{2 P_1} = \frac{2\sqrt{2}}{2} = \sqrt{2} : 1$.

Similar Questions

What is the value of linear velocity,if $\vec \omega = 3\hat i - 4\hat j + \hat k$ and $\vec r = 5\hat i - 6\hat j + 6\hat k$?

If the pair of lines $ax^2-7xy-3y^2=0$ and $2x^2+xy-6y^2=0$ have exactly one line in common and $a$ is an integer,then the equation of the pair of bisectors of the angles between the lines $ax^2-7xy-3y^2=0$ is

The biquadratic equation,two of whose roots are $1+i$ and $1-\sqrt{2}$,is

Electrons with de-Broglie wavelength $\lambda$ fall on the target of an $X$-ray tube. The cut-off wavelength of the emitted $X$-ray is:

The anode voltage of a photocell is kept fixed. The wavelength $\lambda$ of the light falling on the cathode is gradually changed. The plate current $I$ of the photocell varies as follows:

Vedclass Test Series

Exam Paper Generator

Online Exam Module