One mole of an ideal gas at an initial temper | vedclass

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

Question

(D) For an adiabatic process,the work done by the gas is given by the formula:$W = \frac{nR(T_i - T_f)}{\gamma - 1}$Given:Number of moles $n = 1$Work

Vedclass · Accepted Answer

$(T-4) \ K$

Vedclass · Answer

(D) For an adiabatic process,the work done by the gas is given by the formula:$W = \frac{nR(T_i - T_f)}{\gamma - 1}$Given:Number of moles $n = 1$Work done $W = 6R$Initial temperature $T_i = T$Ratio of specific heats $\gamma = 5/3$Substituting these values into the formula:$6R = \frac{1 \cdot R(T - T_f)}{(5/3) - 1}$$6R = \frac{R(T - T_f)}{2/3}$$6R = \frac{3R(T - T_f)}{2}$Dividing both sides by $R$:$6 = \frac{3(T - T_f)}{2}$$12 = 3(T - T_f)$$4 = T - T_f$$T_f = T - 4$Therefore,the final temperature of the gas is $(T - 4) \ K$.

One mole of an ideal gas at an initial temperature of $T$ $K$ does $6R$ of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5/3$,the final temperature of the gas will be $(R=8.31 \ J \ mole^{-1} \ K^{-1})$.

Similar Questions

Is it possible to increase the temperature of a gas without adding heat to it? Explain.

$A$ fixed amount of a gas undergoes a thermodynamic process as shown in the figure,such that the heat interaction along path $B \to C \to A$ is equal to the work done by the gas along path $A \to B \to C$. Then the process $A \to B$ is:

The temperature of the system decreases in the process of

An ideal gas at atmospheric pressure is adiabatically compressed so that its density becomes $32$ times its initial value. If the final pressure of the gas is $128$ atmospheres,the value of $\gamma$ for the gas is:

In an adiabatic change,the pressure $P$ and temperature $T$ of a diatomic gas are related by the relation $P \propto T^C$,where $C$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module