A perfect gas of a given mass is heated first | vedclass

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

Question

(D) According to the ideal gas equation:$P V = n R T$Since the volume $V$ is constant for each vessel,we can write:$P = \left( \frac{n R}{V} \right) T

Vedclass · Accepted Answer

linear with different slopes

Vedclass · Answer

(D) According to the ideal gas equation:$P V = n R T$Since the volume $V$ is constant for each vessel,we can write:$P = \left( \frac{n R}{V} \right) T$This equation is of the form $y = m x$,where $y = P$,$x = T$,and the slope $m = \frac{n R}{V}$.Since the mass is given,$n$ is constant. However,the volumes $V$ of the small and large vessels are different.Therefore,the slope $m = \frac{n R}{V}$ will be different for the two vessels.Thus,the $P-T$ curves are linear,passing through the origin,but with different slopes.

$A$ perfect gas of a given mass is heated first in a small vessel and then in a large vessel,such that their volumes remain unchanged. The $P-T$ curves are

Similar Questions

$A$ gas is enclosed in a vessel at a constant temperature at a pressure of $5 \text{ atm}$ and volume $4 \text{ L}$. Due to a leakage in the vessel,after some time,the pressure is reduced to $4 \text{ atm}$. As a result,the:

State Gay-Lussac's Law of Gaseous Volumes.

The figure shows the $P-V$ diagram of two different masses $m_1$ and $m_2$ of the same gas drawn at a constant temperature $T$. Then:

$A$ balloon contains $500 \ m^3$ of helium at $27^{\circ}C$ and $1 \ atm$ pressure. The volume of the helium at $-3^{\circ}C$ temperature and $0.5 \ atm$ pressure will be ..... $m^3$.

When a gas in a closed vessel was heated so as to increase its temperature by $5^{\circ}C$,there occurred an increase of $1\%$ in its pressure. The original temperature of the gas was ...... $^{\circ}C$

Vedclass Test Series

Exam Paper Generator

Online Exam Module