If 100 L of gas is enclosed in a cylinder, a | vedclass

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

Question

(D) Given: $q = 302.6 \ J$ (heat absorbed by the system is positive). $P_{ext} = 2 \ Pa$. $V_1 = 100 \ L$, $V_2 = 200 \ L$. $\Delta V = V_2 - V_1 = 20

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(D) Given: $q = 302.6 \ J$ (heat absorbed by the system is positive). $P_{ext} = 2 \ Pa$. $V_1 = 100 \ L$, $V_2 = 200 \ L$. $\Delta V = V_2 - V_1 = 200 \ L - 100 \ L = 100 \ L$. Since $1 \ L \cdot Pa = 1 \ J$ is not the standard conversion (usually $1 \ L \cdot atm = 101.3 \ J$), we use the given units directly: Work done $W = -P_{ext} \Delta V = -(2 \ Pa) 	imes (100 \ L) = -200 \ J$. According to the First Law of Thermodynamics: $\Delta U = q + W$. $\Delta U = 302.6 \ J + (-200 \ J) = 102.6 \ J$. The closest value among the options is $100 \ J$.

If $100 \ L$ of gas is enclosed in a cylinder, absorbs $302.6 \ J$ of heat and expands to $200 \ L$ against a constant external pressure of $2 \ Pa$, calculate the internal energy change of the gas. (in $J$)

Similar Questions

Calculate the internal energy change of one mole of an ideal gas during an adiabatic free expansion process from a volume of $2 \ L$ to $5 \ L$.

Joule-Thomson expansion is:

An isolated chamber is divided into two halves by a partition,with an ideal gas in one half. By making a hole in the partition,the gas is allowed to expand to the full chamber. Among the following,the parameter which changes in the process is

During the free expansion of an ideal gas,its.......

In an isobaric process,the ratio of heat supplied to the system $(dQ)$ and work done by the system $(dW)$ for a diatomic gas is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module